# Definite Integration

This section covers:

# Introduction to Definite Integrals

Up to now, we’ve studied the indefinite integral, which is just the function that you get when you integrate another function.

The definite integral is actually a number that represents the area under the curve of that function from an “$$x$$” position to another “$$x$$” position (we just learned how to get this area using Riemann Sums).

It’s pretty crazy that the integral is an area under a curve, but it helps when you think of the equation $$\text{Distance}=\text{Rate}\,(\text{Velocity})\times \text{Time}$$: for the case when the $$x$$-axis of the curve represents time, and $$y$$-axis represents rate, the area (length times width) can represent a distance (or change in position).

We can use this principle to determine how much something changes (for example, its distance) over time.

Here is the Definite Integral as the Area of a Region:

Definite Integral as the Area of a Region Let $$f$$ be continuous and above the $$y$$-axis (non-negative) on interval $$[a,b]$$. The area of the region bounded by $$f$$, the $$x$$-axis and vertical lines at $$x=a$$ and $$x=b$$ (lower and upper limits) is:

$$\displaystyle \text{Area}=\int\limits_{a}^{b}{{f\left( x \right)}}\,dx$$

We’ll see soon that to get this area, we take the integral of $$f\left( x \right)$$, plug $$b$$ in for $$x$$ and then subtract from that value what we get by plugging in $$a$$ for $$x$$.

Note:  The area of the region represented by an integral is only applicable if the region in the interval is totally above the $$\boldsymbol{x}$$-axis (positive $$y$$). We’ll learn later that if any part of the graph is below the $$x$$-axis (negative $$y$$), to get that definite integral, we’ll take the “negative of the area”.

(Thus, to get the definite integral of a function that is both above and below the $$x$$-axis, we can subtract the area above the $$x$$-axis by the area below the $$x$$-axis in that interval.)

# Properties of Definite Integrals

Definite Integrals have some properties; think of these properties just like the properties of any type of area. Most are somewhat obvious:

Properties of Definite Integrals

• $$\displaystyle \int\limits_{a}^{a}{{f\left( x \right)}}\,dx=0$$, if the function is defined at $$x=a$$ (If you stay at one point, you don’t have any area.)
• $$\displaystyle \int\limits_{b}^{a}{{f\left( x \right)}}\,dx=-\int\limits_{a}^{b}{{f\left( x \right)}}\,dx$$, if the function is integrable on $$[a,b]$$. (Think of going backwards and “erasing” area.)
• $$\displaystyle \int\limits_{c}^{a}{{f\left( x \right)}}\,dx=\int\limits_{a}^{b}{{f\left( x \right)}}\,dx+\int\limits_{b}^{c}{{f\left( x \right)}}\,dx$$, if the function is integrable on $$[a,c]$$. (You can add areas.)
• $$\displaystyle \int\limits_{a}^{b}{{k\cdot f\left( x \right)}}\,dx=k\cdot \int\limits_{a}^{b}{{f\left( x \right)}}\,dx$$, if the function is integrable on $$[a,b]$$. (We can move scalars to outside of areas/integrals.)
• $$\displaystyle \int\limits_{a}^{b}{{f\left( x \right)}}\,\pm g\left( x \right)dx=\int\limits_{a}^{b}{{f\left( x \right)}}\,dx\,\,\pm \,\,\int\limits_{a}^{b}{{g\left( x \right)}}\,dx$$, if the function is integrable on $$[a,b]$$. (We can split up areas/integrals).

Now let’s do some problems that demonstrate the definite integral as an area:

 Definite Integral Problem Solution Definite Integral Problem Solution Set up a definite integral that yields the following area: $$f\left( x \right)=4$$ $$\displaystyle f\left( x \right)=\int\limits_{0}^{3}{{4\,dx}}$$   (We can see that the area is $$b\cdot h=\left( 3 \right)\left( 4 \right)=12$$ ) Set up a definite integral that yields the following area: $$f\left( x \right)={{x}^{2}}+3$$ $$\displaystyle f\left( x \right)=\int\limits_{{.5}}^{2}{{\left( {{{x}^{2}}+3} \right)\,dx}}$$   (We can’t easily get the area geometrically) Sketch a graph whose area is given by the definite integral.   Then use a geometric formula to find evaluate the integral. $$\displaystyle \int\limits_{0}^{3}{{x\,dx}}$$ Area (Triangle) = $$\displaystyle \frac{1}{2}bh=\frac{1}{2}\left( 3 \right)\left( 3 \right)=4.5$$ Sketch a graph whose area is given by the definite integral.   Then use a geometric formula to find evaluate the integral. $$\displaystyle \int\limits_{{-3}}^{3}{{\left( {3-\left| x \right|} \right)\,dx}}$$ Area (Triangle) = $$\displaystyle \frac{1}{2}bh=\frac{1}{2}\left( 6 \right)\left( 3 \right)=18$$

Here are a few problems that illustrate the properties of definite integrals. Note that not all of these integrals may be areas, since some are negative (we’ll soon learn that if part of the function is under the $$\boldsymbol{x}$$-axis, the integral is a negative “area” – thus not really an area, but we can use this “area” and make it negative).

 Definite Integral Problem Solution Evaluate the integrals, given the following values:   $$\displaystyle \int\limits_{3}^{5}{{dx}}=4\,\,\,\,\,\,\,\,\,\int\limits_{3}^{5}{{x\,dx}}=10\,\,\,\,\,\,\,\,\,\int\limits_{3}^{5}{{{{x}^{2}}dx}}=50$$     a. $$\displaystyle \int\limits_{5}^{3}{{x\,dx}}$$       b. $$\displaystyle \int\limits_{3}^{5}{{4x\,dx}}$$      c. $$\displaystyle \int\limits_{3}^{5}{{\left( {\frac{1}{2}{{x}^{2}}+3x-8} \right)dx}}$$ a. $$\displaystyle \int\limits_{5}^{3}{{x\,dx}}=-\int\limits_{3}^{5}{{x\,dx}}=-10$$ b. $$\displaystyle \int\limits_{3}^{5}{{4x\,dx}}=4\int\limits_{3}^{5}{{x\,dx}}=\left( 4 \right)\left( {10} \right)=40$$ c. \displaystyle \begin{align}\int\limits_{3}^{5}{{\left( {\frac{1}{2}{{x}^{2}}+3x-8} \right)\,dx}}&=\frac{1}{2}\int\limits_{3}^{5}{{{{x}^{2}}dx+3\int\limits_{3}^{5}{{x\,dx-8\int\limits_{3}^{5}{d}}}x}}\\&=\frac{1}{2}\left( {50} \right)+3\left( {10} \right)-8\left( 4 \right)=23\end{align} Evaluate the integrals, given the following values:   $$\displaystyle \int\limits_{1}^{4}{{f\left( x \right)dx}}=5\,\,\,\,\,\,\,\,\int\limits_{4}^{8}{{f\left( x \right)dx}}=-3\,\,\,\,\,\,\,\,\int\limits_{4}^{8}{{g\left( x \right)dx}}=15$$      a. $$\displaystyle \int\limits_{1}^{8}{{f\left( x \right)\,dx}}$$       b. $$\displaystyle \int\limits_{4}^{1}{{-4f\left( x \right)\,dx}}$$    c. $$\displaystyle \int\limits_{4}^{8}{{\left[ {3f\left( x \right)-g\left( x \right)} \right]\,dx}}$$       d. $$\displaystyle \int\limits_{4}^{4}{{4f\left( x \right)\,dx}}$$ a. $$\displaystyle \int\limits_{1}^{8}{{f\left( x \right)\,dx}}=\int\limits_{1}^{4}{{f\left( x \right)\,dx}}+\int\limits_{4}^{8}{{f\left( x \right)\,dx}}=5+-3=2$$ b. \displaystyle \begin{align}\int\limits_{4}^{1}{{-4f\left( x \right)\,dx}}&=-4\int\limits_{4}^{1}{{f\left( x \right)x\,dx}}=-4\left( {-\int\limits_{1}^{4}{{f\left( x \right)x\,dx}}} \right)\\\,\,&=\left( {-4} \right)\left( {-5} \right)=20\end{align} c. \displaystyle \begin{align}\int\limits_{4}^{8}{{\left[ {3f\left( x \right)-g\left( x \right)} \right]\,dx}}&=3\int\limits_{4}^{8}{{f\left( x \right)-\int\limits_{4}^{8}{{g\left( x \right)}}}}\\\,\,&=3\left( {-3} \right)-15=-24\end{align} d.  $$\displaystyle \int\limits_{4}^{4}{{4f\left( x \right)\,dx}}=4\int\limits_{4}^{4}{{f\left( x \right)\,dx}}=4\left( 0 \right)=0$$

# Fundamental Theorem of Calculus

Wow! This sounds important, doesn’t it? That’s because the Fundamental Theorem of Calculus is important; this theorem is used around the world every day to obtain areas (among other things) of all sort of objects. And the great thing about this theorem is it’s so simple to use (especially compared to some of the summing techniques we’ve used). This theorem is also called the “Net Change” Theorem. So here goes:

<

Fundamental Theorem of Calculus

If a function $$f\left( x \right)$$ is continuous on a closed interval $$[a,b]$$, and $$F$$ is an indefinite integral (antiderivative) on that same interval, then:

$$\displaystyle \int\limits_{a}^{b}{{f\left( x \right)}}\,dx=F\left( b \right)-F\left( a \right)$$

For example, to evaluate $$\displaystyle \int\limits_{1}^{2}{{{{x}^{4}}dx}}$$:            $$\displaystyle \int\limits_{1}^{2}{{{{x}^{4}}}}\,dx=\left[ {\frac{{{{x}^{5}}}}{5}} \right]_{1}^{2}=\frac{{{{{\left( 2 \right)}}^{5}}}}{5}-\frac{{{{{\left( 1 \right)}}^{5}}}}{5}=\frac{{32}}{5}-\frac{1}{5}=\frac{{31}}{5}$$

Note that if the function is totally above the $$y$$-axis in the given interval, this calculation is the area between the function and the $$x$$-axis. If the function is totally below the $$x$$-axis in this interval, this calculation is the opposite (negative) of the area between the $$x$$-axis and the function. If this function is both above and below the $$x$$-axis, then this calculation is the area above the $$x$$-axis subtracted by the area below the $$x$$-axis.

## Definite Integrals on the Graphing Calculator

You can evaluate definite integrals in the graphing calculator using the fnInt(, much like you used the nDeriv( for derivatives.

Hit MATH and then scroll down to fnInt( (or hit 9). Put the lower and upper values for the interval and type in the function using the X,T,θ,n key, hitting the right arrow key in between each entry. Then put “$$x$$” after the “$$d$$” for “$$dx$$”, using right arrow key again. (In this case we had to use the arrow key twice, since we had to use it after the exponent to go back down.) You can also go back with the left arrow key if you need to make any changes. Then hit “Enter”: Note that if you have graphed a function in Y=, you can also use 2nd trace (calc) 7 ($$\int{{f\left( x \right)dx}}$$) to integrate the function, and you will be asked to enter lower and upper limits.

Here are some  Definite Integration problems.

Notice that when we are taking the definite integral of an absolute value function,we need to split the function at the points where the absolute values equals 0, and then,as we did in the Piecewise Functions section, either use the original function, or negate the function, depending on the sign of the function (without the absolute value) in that interval.

Note that you can check these using fnInt (MATH 9) on your graphing calculator.

 Definite Integral Problem Solution $$\displaystyle \int\limits_{1}^{3}{{5x\,dx}}$$ $$\displaystyle \int\limits_{1}^{3}{{5x\,dx}}=\left[ {\frac{{5{{x}^{2}}}}{2}} \right]_{1}^{3}=\frac{{5{{{\left( 3 \right)}}^{2}}}}{2}-\frac{{5{{{\left( 1 \right)}}^{2}}}}{2}=\frac{{45}}{2}-\frac{5}{2}=20$$ $$\displaystyle \int\limits_{0}^{8}{{\left( {\sqrt{t}-4} \right)\,dx}}$$ \displaystyle \begin{align}\int\limits_{0}^{8}{{\left( {{{t}^{{\frac{1}{3}}}}-4} \right)\,dt}}&=\left[ {\frac{{{{t}^{{\frac{4}{3}}}}}}{{\frac{4}{3}}}-4t} \right]_{0}^{8}=\left[ {\frac{{3{{t}^{{\frac{4}{3}}}}}}{4}-4t} \right]_{0}^{8}\\&=\left( {\frac{{3{{{\left( 8 \right)}}^{{\frac{4}{3}}}}}}{4}-4\left( 8 \right)} \right)-\left( {\frac{{3{{{\left( 0 \right)}}^{{\frac{4}{3}}}}}}{4}-4\left( 0 \right)} \right)=-20\end{align} $$\displaystyle \int\limits_{0}^{{\frac{\pi }{2}}}{{\frac{{\cot x-\cos x}}{{\cot x}}\,dx}}$$ \displaystyle \begin{align}\int\limits_{0}^{{\frac{\pi }{2}}}{{\frac{{\cot x-\cos x}}{{\cot x}}\,dx}}&=\int\limits_{0}^{{\frac{\pi }{2}}}{{1-\frac{{\cos x}}{{\left( {\frac{{\cos x}}{{\sin x}}} \right)}}\,=\int\limits_{0}^{{\frac{\pi }{2}}}{{1-\sin x\,dx}}=\left[ {x+\cos x} \right]_{0}^{{\frac{\pi }{2}}}dx}}\\\,\,\,\,\,&=\left[ {\frac{\pi }{2}+\cos \left( {\frac{\pi }{2}} \right)} \right]-\left[ {0+\cos \left( 0 \right)} \right]=\left( {\frac{\pi }{2}+0} \right)-\left( {0+1} \right)\\&=\frac{\pi }{2}-1\approx .571\,\,\,\,\,\end{align} $$\displaystyle \int\limits_{{-5}}^{5}{{\left| {{{x}^{2}}-4} \right|\,dx}}$$ What graph looks like: Note that we can integrate with a calculator: First, create a sign chart to see where the function is positive or negative: For the positive intervals, we just use $${{x}^{2}}-4$$ as is. Where it is negative, we need to negate $${{x}^{2}}-4$$. This is because the absolute value function only returns a positive function, so when we remove the absolute value to integrate, we need to adjust the underlying function. Separate into three integrals for these intervals: \displaystyle \begin{align}\int\limits_{{-5}}^{5}{{\left| {{{x}^{2}}-4} \right|\,dx}}&=\int\limits_{{-5}}^{{-2}}{{\left( {{{x}^{2}}-4} \right)\,dx+}}\int\limits_{{-2}}^{2}{{\left( {-{{x}^{2}}+4} \right)\,dx}}+\int\limits_{2}^{5}{{\left( {{{x}^{2}}-4} \right)\,dx}}\\&=\left[ {\left( {\frac{{{{x}^{3}}}}{3}-4x} \right)} \right]_{{-5}}^{{-2}}+\left[ {\left( {-\frac{{{{x}^{3}}}}{3}+4x} \right)} \right]_{{-2}}^{2}+\left[ {\left( {\frac{{{{x}^{3}}}}{3}-4x} \right)} \right]_{2}^{5}\\\,\,&=\left[ {\left( {\frac{{{{{\left( {-2} \right)}}^{3}}}}{3}-4\left( {-2} \right)} \right)-\left( {\frac{{{{{\left( {-5} \right)}}^{3}}}}{3}-4\left( {-5} \right)} \right)} \right]\\&\,\,\,\,\,\,\,\,+\left[ {\left( {-\frac{{{{{\left( 2 \right)}}^{3}}}}{3}+4\left( 2 \right)} \right)-\left( {-\frac{{{{{\left( {-2} \right)}}^{3}}}}{3}+4\left( {-2} \right)} \right)} \right]\\\,\,&\,\,\,\,\,\,\,\,+\left[ {\left( {\frac{{{{{\left( 5 \right)}}^{3}}}}{3}-4\left( 5 \right)} \right)-\left( {\frac{{{{{\left( 2 \right)}}^{3}}}}{3}-4\left( 2 \right)} \right)} \right]\\\,\,\,&=27+\frac{{32}}{3}+27=\frac{{194}}{3}=64\frac{2}{3}\end{align}

And, again, we can use the definite integral to get an area, if the $$y$$ values in the interval are greater than 0 (the function is completely above the $$x$$-axis). Note in the second problem, we have to solve for the $$x$$-intercepts, or zeros, and sketch a graph (or use a sign chart) to see where the function lies above the $$x$$-axis.

 Problem Solution Find the area of the region with the following boundaries: $$f\left( x \right)=3x-5,\,\,y=0,\,\,x=2,\,\,x=4$$ Since this region is all above the $$\boldsymbol{x}$$-axis, we can use the definite integral to get the area: \displaystyle \begin{align}\int\limits_{2}^{4}{{\left( {3x-5} \right)\,dx}}&=\left[ {\frac{3}{2}{{x}^{2}}-5x} \right]_{2}^{4}\\&=\left[ {\frac{3}{2}{{{\left( 4 \right)}}^{2}}-5\left( 4 \right)} \right]-\left[ {\frac{3}{2}{{{\left( 2 \right)}}^{2}}-5\left( 2 \right)} \right]\\&=4-\left( {-4} \right)=8\end{align} Find the area of the region with the following boundaries: $$f\left( x \right)=3\sqrt{x}-x,\,\,\,y=0$$ First, find the zeros to get the interval where the function is above the $$x$$-axis: $$\begin{array}{l}3\sqrt{x}-x=0;\,\,\,\,\,\,3\sqrt{x}=x;\,\,\,\,\,{{\left( {3\sqrt{x}} \right)}^{2}}={{x}^{2}};\,\,\,\,\,\,\,9x={{x}^{2}};\\\,\,\,\,\,{{x}^{2}}-9x=0;\,\,\,\,x\left( {x-9} \right)=0;\,\,\,\,\,\,\,x=0\text{ and }x=9\end{array}$$ Thus, we need $$\displaystyle \int\limits_{0}^{9}{{\left( {3\sqrt{x}-x} \right)\,dx}}$$: \displaystyle \begin{align}\int\limits_{0}^{9}{{\left( {3\sqrt{x}-x} \right)dx}}&=\int\limits_{0}^{9}{{\left( {3{{x}^{{\frac{1}{2}}}}-x} \right)dx}}=\left[ {3\cdot \frac{{{{x}^{{\frac{3}{2}}}}}}{{\left( {\frac{3}{2}} \right)}}-\frac{{{{x}^{2}}}}{2}} \right]_{0}^{9}\\&=\left[ {3\cdot \frac{2}{3}{{{\left( 9 \right)}}^{{\frac{3}{2}}}}-\frac{{{{9}^{2}}}}{2}} \right]-\left[ {3\cdot \frac{3}{2}{{{\left( 0 \right)}}^{{\frac{3}{2}}}}-\frac{{{{0}^{2}}}}{2}} \right]\\&=\left( {54-\frac{{81}}{2}} \right)-0=13.5\end{align}

# Definite Integration and Area

You’ve probably realized by now (and I’ve hinted at it a few times) that to get the value of an integral under the $$x$$-axis, you can take the opposite (negative) of the area of that region. So, you have to really be careful if a problem calls for an integral of a region that is both above and below the $$x$$-axis, you have to basically add up the area above the $$x$$-axis and subtract the area below the $$x$$-axis.

But let’s think about it. If you were to take the absolute value of the function (so that everything moves above the $$x$$-axis), you would have the area!  Here’s an example:

Definite Integrals and Area

Problem:  Below is the graph of a function $$f$$, which consists of straight lines, a semi-circle, and a quarter circle.

Evaluate:

a. $$\displaystyle \int\limits_{{-8}}^{9}{{f\left( x \right)dx}}$$      b. $$\displaystyle \int\limits_{{9}}^{5}{{f\left( x \right)dx}}$$      c. $$\displaystyle \int\limits_{{-8}}^{9}{{\left| {f\left( x \right)} \right|dx}}$$      d. $$\displaystyle \int\limits_{0}^{3}{{\left( {f\left( x \right)+2} \right)dx}}$$      e. $$\displaystyle \int\limits_{2}^{4}{{f\left( {x-1} \right)dx}}$$      f. $$\displaystyle \int\limits_{{-8}}^{9}{{f\left( {\left| x \right|} \right)dx}}$$     g. If $$F\left( {-8} \right)=6$$ and $$F$$ is the antiderivative of $$f$$, find $$F\left( 9 \right)$$. Solution:  Let’s use Geometry to figure out the area in each section; we’ve annotated the graph above using rectangle, triangle, and circle area formulas.

a. $$\displaystyle \int\limits_{{-8}}^{9}{{f\left( x \right)dx}}$$:    Add up the areas in all the 9 regions, using negatives for the areas below the $$x$$-axis:

$$-2+2+8+6.28+6+2+\left( {-.5} \right)+\left( {-3} \right)+\left( {-7.07} \right)=11.71$$

b. $$\displaystyle \int\limits_{9}^{5}{{f\left( x \right)dx}}$$:    Add up the areas between $$x=5$$ and $$x=9$$, but since we’re going “backwards” (from 9 to 5), we have to take the opposite (negative):

$$\left( {-.5} \right)+\left( {-3} \right)+\left( {-7.07} \right)=-10.57\,\,\,\,\,\,\,\,\,-\left( {-10.57} \right)=10.57$$

c. $$\displaystyle \int\limits_{{-8}}^{9}{{\left| {f\left( x \right)} \right|dx}}$$     Add up the absolute value of the areas in all the 9 regions (so disregard the negatives):

$$2+2+8+6.28+6+2+.5+3+7.07=38.85$$

d.  $$\displaystyle \int\limits_{0}^{3}{{\left( {f\left( x \right)+2} \right)dx}}$$    (Vertical Shift Up) Take the area between $$x=0$$ and $$x=3$$ (area is 6), and then imagine the top of it (the $$y$$) is translated up 2 to 4.  Thus, the area would be $$6+6=12$$.

e. $$\displaystyle \int\limits_{2}^{4}{{f\left( {x-1} \right)dx}}$$    (Horizontal Shift to Right) The whole graph is shifted to the right by 1, so to get back to the original graph, we’d have to subtract 1 from the upper and lower bounds:  $$\displaystyle \int\limits_{2}^{4}{{f\left( {x-1} \right)dx}}=\int\limits_{1}^{3}{{f\left( x \right)dx}}=4$$

f.  $$\displaystyle \int\limits_{{-8}}^{9}{{f\left( {\left| x \right|} \right)dx}}$$ (absolute value of the $$x$$): “Erase” what’s to the left of the $$y$$-axis (the negative $$x$$ values) and reflect what’s on the right (positive $$x$$ values), so the function is even. Double the area of the positive $$x$$ (negating the areas below the $$x$$-axis), so we have:

$$2\left[ {6+2+\left( {-.5} \right)+\left( {-3} \right)+\left( {-7.07} \right)} \right]=-5.14$$.

g.  This one’s a little tricky; we have to use the equation $$\displaystyle \int\limits_{{-8}}^{9}{{f\left( x \right)dx}}=F\left( 9 \right)-F\left( {-8} \right)$$.  Thus, $$\displaystyle F\left( 9 \right)=\int\limits_{{-8}}^{9}{{f\left( x \right)dx}}+F\left( {-8} \right)=11.71+6=17.71$$.

Here are a few more problems on Definite Integration and Area:

Definite Integral Properties

ProblemsSupposed that $$f$$ and $$g$$ are continuous functions defined on the interval $$[1, 4]$$:

$$\displaystyle \int\limits_{1}^{4}{{f\left( x \right)dx=5}}$$ and $$\int\limits_{1}^{4}{{g\left( x \right)dx=-3}}$$

Evaluate:

a. $$\displaystyle \int\limits_{1}^{4}{{f\left( x \right)dx=5}}$$, where $$h\left( x \right)=f\left( x \right)+2$$         b. $$\displaystyle \int\limits_{{-1}}^{2}{{k\left( x \right)dx}}$$, where $$k\left( x \right)=f\left( {x+2} \right)$$

c. $$\displaystyle \int\limits_{2}^{5}{{j\left( x \right)dx}}$$, where $$j\left( x \right)=4g\left( x \right)$$                        d. $$\displaystyle \int\limits_{1}^{4}{{\left( {2f-g} \right)dx}}$$

Solutions:

a. (Vertical Shift Up) Since and we are shifting the function vertically up 2, we will add another “block” of area that is 2 units high by 3 units wide. The new area is $$5+2\cdot 3=11$$ (see diagram). This makes sense since

$$\displaystyle \int\limits_{1}^{4}{{\left( {f\left( x \right)+2} \right)}}=\int\limits_{1}^{4}{{f\left( x \right)}}+\int\limits_{1}^{4}{2}=\int\limits_{1}^{4}{{f\left( x \right)}}+\left. {2x} \right|_{1}^{4}=5+\left( {8-2} \right)=11$$.

b. (Horizontal Shift to Left) For $$f\left( {x+2} \right)$$, the function $$f\left( x \right)$$ is shifted to the left by 2, so for $$\int\limits_{{-1}}^{2}{{f\left( {x+2} \right)dx}}$$, we can add 2 to the upper and lower limits and use $$f\left( x \right)$$. Thus, $$\int\limits_{{-1}}^{2}{{f\left( {x+2} \right)dx}}=\int\limits_{1}^{4}{{f\left( x \right)dx}}=5$$.

c. Since we have a scalar out in front of the $$g\left( x \right)$$, we’ll just multiply the integral amount by that scalar:

$$\displaystyle \int\limits_{1}^{4}{{\left[ {4g\left( x \right)} \right]dx=4\int\limits_{1}^{4}{{g\left( x \right)dx=4\cdot -3}}}}=-12$$.

d. We can add and subtract integrals, as well as multiply them by scalars:

$$\displaystyle \int\limits_{1}^{4}{{\left( {2f-g} \right)dx}}=\int\limits_{1}^{4}{{\left[ {2f\left( x \right)} \right]dx-\int\limits_{1}^{4}{{g\left( x \right)dx=2\int\limits_{1}^{4}{{f\left( x \right)dx-\int\limits_{1}^{4}{{g\left( x \right)dx=2\cdot 5-}}}}\left( {-3} \right)}}}}=13$$.

Definite Integral Properties

Find $$\displaystyle \int\limits_{2}^{6}{{f\left( x \right)dx}}$$, if:   a. $$f\left( x \right)$$ is even, $$\displaystyle \int\limits_{{-2}}^{2}{{f\left( x \right)dx}}=8$$ and $$\displaystyle \int\limits_{{-6}}^{6}{{f\left( x \right)dx}}=20$$     b. $$f\left( x \right)$$ is odd, and $$\displaystyle \int\limits_{{-2}}^{6}{{f\left( x \right)dx=8}}$$     c. $$\displaystyle \int\limits_{2}^{6}{{\left( {2f\left( x \right)-1} \right)dx}}=14$$

Solutions:

a. The best way to do these types of problems is to draw a picture. An even function is symmetrical around the $$y$$-axis, meaning any area on the right-hand side (positive $$x$$) is the same as that on the left-hand side (negative $$x$$), for the same distance from the origin. Divide up the areas given ($$4+4=8$$ and $$10+10=20$$) to get the area in the interval $$[2,6]$$ to be $$6$$. b. Again, draw a picture. Since the function is odd, it’s symmetrical around the origin, meaning that any area on the right-hand side (positive $$x$$) is the opposite (negative) of that on the left-hand side (negative $$x$$), for the same distance from the origin. Thus, the areas in the interval $$[-2,0]$$ and $$[0,2]$$ cancel out (add up to 0). This makes the area in the interval $$[2,6]$$ the same as the area in the interval $$[-2,6]$$, which is $$8$$. c. We can break up the integral to solve back for $$\displaystyle \int\limits_{2}^{6}{{f\left( x \right)dx}}$$:

\displaystyle \begin{align}14&=\int\limits_{2}^{6}{{\left( {2f\left( x \right)-1} \right)dx}}=2\int\limits_{2}^{6}{{f\left( x \right)dx-\int\limits_{2}^{6}{{1dx}}}}=2\int\limits_{2}^{6}{{f\left( x \right)dx-\left. x \right|_{{x=2}}^{{x=6}}}}\\14&=2\int\limits_{2}^{6}{{f\left( x \right)dx-\left( {6-2} \right)}}=2\int\limits_{2}^{6}{{f\left( x \right)dx-4}}\\14&=2\int\limits_{2}^{6}{{f\left( x \right)dx-4}};\,\,\,\,\,\,18=2\int\limits_{2}^{6}{{f\left( x \right)dx;\,\,\,\,}}\int\limits_{2}^{6}{{f\left( x \right)dx=9\,\,}}\end{align}

Definite Integral Trig Problems

Problems:  Evaluate the following (using geometry), given the fact that :

a. $$\displaystyle \int\limits_{0}^{{2\pi }}{{\sin \left( x \right)dx}}$$      b. $$\displaystyle \int\limits_{0}^{{\frac{\pi }{2}}}{{\sin \left( x \right)dx}}$$      c. $$\displaystyle \int\limits_{0}^{{2\pi }}{{\left| {\sin x} \right|\,dx}}$$       d. $$\displaystyle \int\limits_{{-\frac{\pi }{2}}}^{{\frac{\pi }{2}}}{{\cos \left( x \right)\,dx}}$$      e. $$\displaystyle \int\limits_{0}^{{\frac{\pi }{2}}}{{\left( {\sin \left( x \right)+2x} \right)\,dx}}$$ Solutions:

a. Since $$\int\limits_{0}^{\pi }{{\sin \left( x \right)dx}}=2$$, $$\int\limits_{0}^{{2\pi }}{{\sin \left( x \right)dx}}$$ is this amount above the $$x$$-axis (2), and also the opposite (negative) of this amount below the $$x$$-axis (–2). Thus, we have $$2+-2=0$$.

b. Since $$\int\limits_{0}^{\pi }{{\sin \left( x \right)dx}}=2$$, $$\int\limits_{0}^{{\frac{\pi }{2}}}{{\sin \left( x \right)dx}}$$ is half this amount, which is 1. (The function is symmetrical across $$\displaystyle x=\frac{\pi }{2}$$).

c. Since $$\int\limits_{0}^{\pi }{{\sin \left( x \right)dx}}=2$$, $$\int\limits_{0}^{{2\pi }}{{\left| {\sin \left( x \right)} \right|dx}}$$ is this amount  above the $$x$$-axis (2, and again above the $$x$$-axis because of the absolute value. Thus, we have $$2+2=4$$. d. (Horizontal Shift by $$\displaystyle \frac{\pi }{2}$$ to Left or Right) The graph of cosine is the graph of sin that is shifted either $$\displaystyle \frac{\pi }{2}$$ to the left or right. Thus, $$\int\limits_{{-\frac{\pi }{2}}}^{{\frac{\pi }{2}}}{{\cos x\,dx}}=\int\limits_{0}^{\pi }{{\sin x\,dx}}$$, which is $$2$$.

e. $$\int\limits_{0}^{{\frac{\pi }{2}}}{{\left( {\sin \left( x \right)+2x} \right)\,dx}}=\int\limits_{0}^{{\frac{\pi }{2}}}{{\sin \left( x \right)\,dx}}+\int\limits_{0}^{{\frac{\pi }{2}}}{{2x\,dx}}$$. We know that $$\int\limits_{0}^{{\frac{\pi }{2}}}{{\sin \left( x \right)\,dx}}=1$$ from b. above. $$\int\limits_{0}^{{\frac{\pi }{2}}}{{2x\,dx}}$$ is a triangle (see diagram) with a base of $$\displaystyle \frac{\pi }{2}$$ and a height of $$\displaystyle 2\cdot \frac{\pi }{2}=\pi$$, so its area is $$\displaystyle \frac{1}{2}bh=\frac{1}{2}\cdot \frac{\pi }{2}\cdot \pi =\frac{{{{\pi }^{2}}}}{4}$$. The total area is $$\displaystyle 1+\frac{{{{\pi }^{2}}}}{4}\approx 3.47$$.

# Mean Value Theorem (MVT) for Integrals

We learned about the Mean Value Theorem for Derivatives here in the Curve Sketching section. The Mean Value Theorem (MVT) for Integrals is a theorem that guarantees that a continuous function in an interval contains at least one point where that function is equal to the average value of the function.

So what does this mean in plain English? All it really means is that for a continuous function between two different points, there is at least one point where the “$$y$$” value is equal to the average of the all the “$$y$$” values in that interval. In terms of geometry, this means that there exists a rectangle between the two points whose area is equal to the area under the curve of the function between those two points. Think of flattening a mountain so it fills the valleys perfectly.

Here’s the formal definition of the Mean Value Theorem for Integrals:

Mean Value Theorem for Integrals For a function $$f$$ that is continuous on closed interval $$[a,b]$$, there exists at least one number $$c$$ in that closed interval such that:

$$\displaystyle \,\,\,\,\,\,\,\,\,\,\,\int\limits_{a}^{b}{{f\left( x \right)\,dx=f\left( c \right)}}\left( {b-a} \right)$$

For a geometric explanation, think of $$f\left( c \right)$$ as the height of a rectangle and $$\left( {b-a} \right)$$ as the width. There is a number $$c$$ such that  has the same area as the region under the curve of that function from $$a$$ to $$b$$.

We’ll use this formula to solve problems where we find the “$$c$$” guaranteed by the Mean Value Theorem for an integral in a specific interval. We’ll also derive the Average Value of a Function from this formula.

# Average Value of a Function

Now we can solve for the Average Value of a Function by dividing both sides of the Mean Value Theorem equation by $$\left( {b-a} \right)$$. The average value of a function is the $$f\left( c \right)$$ in the Mean Value Theorem equation.

Average Value of a Function

If a function $$f$$ is integrable on a closed interval $$[a,b]$$, then the average value on that interval is:

$$\displaystyle \frac{1}{{\left( {b-a} \right)}}\int\limits_{a}^{b}{{f\left( x \right)\,dx}}$$

To remember this, think Integral over Interval:

$$\displaystyle \text{Average Value}=\frac{{\int\limits_{a}^{b}{{f\left( x \right)\,dx}}}}{{\left( {b-a} \right)}}=\,\,\frac{{\text{Integral}}}{{\text{Interval}}}$$

Here are the types of problems you might see for the Mean Value Theorem:

 Mean Value Theorem Problem Solution Find the value of “ $$c$$” that is guaranteed by the Mean Value Theorem (MVT) for integrals for $$f\left( x \right)=-{{x}^{2}}-8$$ in the interval $$\left[ {1,4} \right]$$. Use the MVT equation to find the “$$c$$”: $$\begin{array}{c}\int\limits_{a}^{b}{{f\left( x \right)\,dx=f\left( c \right)}}\left( {b-a} \right):\,\,\,\,\,\,\int\limits_{1}^{4}{{\left( {-{{x}^{2}}-8} \right)\,dx=f\left( c \right)}}\left( {4-1} \right)\\\left[ {-\frac{{{{x}^{3}}}}{3}-8x} \right]_{1}^{4}=\left( {-{{c}^{2}}-8} \right)\left( 3 \right)\\\left( {-\frac{{{{4}^{3}}}}{3}-8\left( 4 \right)} \right)-\left( {-{{{\frac{1}{3}}}^{3}}-8\left( 1 \right)} \right)=3\left( {-{{c}^{2}}-8} \right)\\-45=-3{{c}^{2}}-24;\,\,\,\,\,{{c}^{2}}=7;\,\,\,\,\,c=\pm \,\sqrt{7}\\\,\,c=\sqrt{7}\end{array}$$ Take positive only since the “$$c$$” need to be in the interval $$\left[ {1,4} \right]$$. Find the value of “$$c$$” that is guaranteed by the Mean Value Theorem (MVT) for integrals for $$f\left( x \right)=3{{\sec }^{2}}\left( x \right)$$ in the interval $$\displaystyle \left[ {-\frac{\pi }{4},\frac{\pi }{4}} \right]$$. Use the MVT equation to find the “$$c$$”: $$\begin{array}{c}\int\limits_{a}^{b}{{f\left( x \right)\,dx=f\left( c \right)}}\left( {b-a} \right):\,\,\int\limits_{{-\frac{\pi }{4}}}^{{\frac{\pi }{4}}}{{\left( {3{{{\sec }}^{2}}x} \right)\,dx=f\left( c \right)}}\left[ {\frac{\pi }{4}-\left( {-\frac{\pi }{4}} \right)} \right]\\\left[ {3\tan x} \right]_{{-\frac{\pi }{4}}}^{{\frac{\pi }{4}}}=\left( {3{{{\sec }}^{2}}c} \right)\left[ {\frac{\pi }{4}-\left( {-\frac{\pi }{4}} \right)} \right]\\\left( {3\tan \left( {\frac{\pi }{4}} \right)} \right)-\left( {3\tan \left( {-\frac{\pi }{4}} \right)} \right)=\left( {3{{{\sec }}^{2}}c} \right)\left( {\frac{\pi }{2}} \right)\\3\cdot 1-\left( {3\cdot -1} \right)=\frac{{3\pi }}{2}{{\sec }^{2}}c;\,\,\,\,\,6=\frac{{3\pi }}{2}{{\sec }^{2}}c;\,\,\,\,\,\,\,{{\sec }^{2}}c=\frac{4}{\pi }\\{{\cos }^{2}}c=\frac{\pi }{4};\,\,\,\,\,\,\,\cos c=\pm \,\sqrt{{\frac{\pi }{4}}};\,\,\,\,\,\,\,\,c=\pm \,\arccos \sqrt{{\frac{\pi }{4}}}\\c=\pm \,.482\end{array}$$ Use both positive and negative, since both values are in the interval $$\displaystyle \left[ {-\frac{\pi }{4},\frac{\pi }{4}} \right]$$. Find the value of “$$c$$” that is guaranteed by the Mean Value Theorem (MVT) for integrals for $$f\left( x \right)=\ln \left( {x+2} \right)$$ in the interval $$\left[ {-1,\,4} \right]$$. (Use calculator to get the integral of the ln function.) Use the MVT equation to find the “$$c$$”: $$\begin{array}{c}\int\limits_{a}^{b}{{f\left( x \right)\,dx=f\left( c \right)}}\left( {b-a} \right):\,\,\,\,\,\,\int\limits_{{-1}}^{4}{{\ln \left( {x+2} \right)\,dx=f\left( c \right)}}\left[ {4-\left( {-1} \right)} \right]\\5.75=\ln \left( {c+2} \right)\left( 5 \right);\,\,\,\,1.15=\ln \left( {c+2} \right);\,\,\,\,\,\,{{e}^{{1.15}}}=c+2\\c\approx \,\,1.158\end{array}$$

Here are some Average Value Theorem problems:

 Average Value Theorem Problem Solution Find the Average Value of $$\displaystyle f\left( x \right)=2+\frac{8}{{{{x}^{2}}}}$$ from $$\left[ {2,4} \right]$$. Think “Integral Over Interval”:      Average Value = $$\displaystyle \frac{1}{{\left( {b-a} \right)}}\int\limits_{a}^{b}{{f\left( x \right)\,dx}}$$ $$\displaystyle \begin{array}{c}\frac{1}{{\left( {4-2} \right)}}\int\limits_{2}^{4}{{\left( {2+\frac{8}{{{{x}^{2}}}}} \right)\,dx}}=\frac{1}{2}\int\limits_{2}^{4}{{\left( {2+8{{x}^{{-2}}}} \right)\,dx}}=\frac{1}{2}\left[ {2x-8{{x}^{{-1}}}} \right]_{2}^{4}\\=\frac{1}{2}\left( {\left[ {2\left( 4 \right)-\frac{8}{4}} \right]-\left[ {2\left( 2 \right)-\frac{8}{2}} \right]} \right)=\frac{1}{2}\left( {8-2-4+4} \right)=3\end{array}$$ Find the Average Value of $$f\left( x \right)=\cos x+3$$ from $$\displaystyle \left[ {-\frac{\pi }{2},\,\frac{\pi }{2}} \right]$$. Think “Integral Over Interval”:      Average Value = $$\displaystyle \frac{1}{{\left( {b-a} \right)}}\int\limits_{a}^{b}{{f\left( x \right)\,dx}}$$ $$\displaystyle \begin{array}{c}\frac{1}{{\left[ {\frac{\pi }{2}-\left( {-\frac{\pi }{2}} \right)} \right]}}\,\int\limits_{{-\frac{\pi }{2}}}^{{\frac{\pi }{2}}}{{\left( {\cos x+3} \right)\,dx}}=\frac{1}{\pi }\int\limits_{{-\frac{\pi }{2}}}^{{\frac{\pi }{2}}}{{\left( {\cos x+3} \right)\,dx}}=\frac{1}{\pi }\left[ {\sin x+3x} \right]_{{-\frac{\pi }{2}}}^{{\,\,\,\frac{\pi }{2}}}\\=\frac{1}{\pi }\left( {\left[ {\sin \left( {\frac{\pi }{2}} \right)+3\left( {\frac{\pi }{2}} \right)} \right]-\left[ {\sin \left( {-\frac{\pi }{2}} \right)+3\left( {-\frac{\pi }{2}} \right)} \right]} \right)\\=\frac{1}{\pi }\left[ {1+\frac{{3\pi }}{2}-\left( {-1} \right)+\frac{{3\pi }}{2}} \right]=\frac{1}{\pi }\left( {2+3\pi } \right)=\,\,\frac{2}{\pi }+3\end{array}$$ Find the number(s) $$b$$ such that the Average Value of $$f\left( x \right)=6x-3$$ on interval $$\left[ {0,b} \right]$$ is 15. Think “Integral Over Interval”:      Average Value = $$\displaystyle \frac{1}{{\left( {b-a} \right)}}\int\limits_{a}^{b}{{f\left( x \right)\,dx}}$$ $$\displaystyle \begin{array}{c}\frac{1}{{b-0}}\int\limits_{0}^{b}{{\left( {6x-3} \right)\,dx}}=\frac{1}{b}\left[ {3{{x}^{2}}-3x} \right]_{0}^{b}\\\frac{1}{b}\left( {\left[ {3{{b}^{2}}-3b} \right]-\left[ {3{{{\left( 0 \right)}}^{2}}-3\left( 0 \right)} \right]} \right)=15;\,\,\,\,\,\frac{1}{b}\left( {3{{b}^{2}}-3b} \right)=15\\3b-3=15;\,\,\,\,3b=18;\,\,\,\,\,b=6\end{array}$$

Here is an application of the Average Value:

 Average Value Theorem Application Solution The temperature over a 10-hour period can be modeled by the function $$f\left( t \right)=-{{t}^{2}}+4t+45$$.   a. Find the average temperature. b. Find the minimum temperature. c. Find the maximum temperature. a. The average value over the 10-hour interval is: $$\displaystyle \frac{1}{{\left( {b-a} \right)}}\int\limits_{a}^{b}{{f\left( t \right)\,dt}}=\frac{1}{{\left( {10-a} \right)}}\int\limits_{0}^{{10}}{{\left( {-{{t}^{2}}+4t+45} \right)dt}}=\left( {\frac{1}{{10}}} \right)316\frac{2}{3}=31\frac{2}{3}$$ degrees. To get the minimum and maximum temperatures over that period, let’s first graph, since we have to check the endpoints, too: b. From the graph above, in the interval $$0\le t\le 10$$, the minimum temperature is –15 degrees, at the endpoint when . c. From the graph above, in the interval $$0\le t\le 10$$, the maximum temperature is 49 degrees.

# 2nd Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus has to do with taking the derivative of the definite integral; we basically “undo” the integral to get the original function back. We have to be careful though, since we don’t always get the original function exactly as it was: it turns out (because of the chain rule), we have to multiply the function by the derivative of the upper limit of the interval. And to use this theorem, the lower bound must be a constant.

This theorem says that $$\displaystyle \frac{d}{{dx}}\left( {\int\limits_{a}^{x}{{f\left( t \right)\,dt}}} \right)=f\left( x \right)$$, or if $$\displaystyle F\left( x \right)=\int\limits_{a}^{x}{{f\left( t \right)\,dt}}$$, then $${F}’\left( x \right)=f\left( x \right)$$.

It looks like the derivative of an integral (accumulation function) gets us back to the original integrand with just a change of variables. But let’s keep going, using the First Fundamental Theorem of Calculus:  $$\displaystyle \frac{d}{{dx}}\left( {\int\limits_{a}^{{g\left( x \right)}}{{f\left( t \right)\,dt}}} \right)=\frac{d}{{dx}}\left( {F\left( {g\left( x \right)} \right)-F\left( a \right)} \right)=\frac{d}{{dx}}F\left( {g\left( x \right)} \right)=f\left( {g\left( x \right)} \right)\cdot {g}’\left( x \right)$$, where $$a$$ is a constant. The $$F\left( a \right)$$ disappears, since the derivative of a constant is 0, but for the $$F\left( x \right)$$, we substitute the upper limit in for the variable, but then have to use the chain rule to multiply by the derivative of this function.

Let’s show an example, where we solve the definite integral and then take the derivative back. Note that you won’t have to do this much work for each problem; we’ll see soon that we just substitute the upper limit in the integral function, and then multiply by the derivative of that upper limit.

Find the derivative of:    $$\displaystyle F\left( x \right)=\int\limits_{\pi }^{{2{{x}^{3}}}}{{\sin \left( t \right)\,dt}}$$:

First solve for the integral:    $$\displaystyle \int\limits_{\pi }^{{2{{x}^{3}}}}{{\sin \left( t \right)\,dt=\left[ {-\cos \left( t \right)} \right]}}_{\pi }^{{2{{x}^{3}}}}=-\cos \left( {2{{x}^{3}}} \right)-\left( {-\cos \left( \pi \right)} \right)=-\cos \left( {2{{x}^{3}}} \right)-1$$

Then take the derivative:    $$\displaystyle \frac{d}{{dx}}\left( {-\cos \left( {2{{x}^{3}}} \right)-1} \right)=\frac{d}{{dx}}\left( {-\cos \left( {2{{x}^{3}}} \right)} \right)-\frac{d}{{dx}}\left( 1 \right)=\sin \left( {2{{x}^{3}}} \right)\cdot 6{{x}^{2}}-0=6{{x}^{2}}\sin \left( {2{{x}^{3}}} \right)$$

Here’s the formal definition of the 2nd Fundamental Theorem of Calculus, with the basic instructions on how to solve these problems:

2nd Fundamental Theorem of Calculus

If $$f$$ is continuous on an open interval that contains $$a$$ (a constant), for every $$x$$ in the interval,

If $$\displaystyle F\left( x \right)=\int\limits_{a}^{x}{{f\left( t \right)\,dt}}$$, then $${F}’\left( x \right)=f\left( x \right)$$            (alternate way: $$\displaystyle \frac{d}{{dx}}\left[ {\int\limits_{a}^{x}{{f\left( t \right)}}F\left( t \right)} \right]=f\left( x \right)$$)

When finding $${F}’\left( x \right)$$, plug in the upper bound into the function $$f\left( t \right)$$ directly, but if the upper bound is different than just plain “$$x$$”, multiply the function by the derivative of the upper bound.

In other words, if $$\displaystyle F\left( x \right)=\int\limits_{a}^{{g\left( x \right)}}{{f\left( t \right)\,dt}}$$, then $${F}’\left( x \right)=f\left( {g\left( x \right)} \right)\cdot {g}’\left( x \right)$$. (For example, if the upper bound is $${{x}^{2}}$$, plug in the $${{x}^{2}}$$ everywhere there’s a $$t$$, but then multiply by $$2x$$, the derivative of $${{x}^{2}}$$).

If there are variables in both the upper and lower bounds, separate the integral into two integrals with constants. For example, separate $$\displaystyle \int\limits_{{f\left( x \right)}}^{{g\left( x \right)}}{{}}$$ into $$\displaystyle \int\limits_{{f\left( x \right)}}^{a}{{}}$$ (which is  $$\displaystyle -\int\limits_{a}^{{f\left( x \right)}}{{}}$$) and $$\displaystyle \int\limits_{a}^{{g\left( x \right)}}{{}}$$.

Here are some Second Theorem of Calculus problems:

 2nd Theorem of Calculus Problem Solution Find the derivative of $$\displaystyle F\left( x \right)=\int\limits_{{-3}}^{x}{{\left( {{{t}^{3}}-4t} \right)\,dt}}$$ $$\displaystyle {F}’\left( x \right)=\left( {{{x}^{3}}-4x} \right)\cdot \frac{d}{{dx}}\left( x \right)=\left( {{{x}^{3}}-4x} \right)\cdot 1={{x}^{3}}-4x$$ Find the derivative of $$\displaystyle F\left( x \right)=\int\limits_{2}^{{{{x}^{4}}}}{{\frac{1}{{{{t}^{2}}}}\,dt}}$$. $$\displaystyle {F}’\left( x \right)=\frac{1}{{{{{\left( {{{x}^{4}}} \right)}}^{2}}}}\cdot \frac{d}{{dx}}\left( {{{x}^{4}}} \right)=\frac{1}{{{{x}^{8}}}}\left( {4{{x}^{3}}} \right)=4{{x}^{{-5}}}=\frac{4}{{{{x}^{5}}}}$$ Note:  Let’s show how this works “the long way”: $$\displaystyle \int\limits_{2}^{{{{x}^{4}}}}{{\frac{1}{{{{t}^{2}}}}\,dt}}=\int\limits_{2}^{{{{x}^{4}}}}{{{{t}^{{-2}}}\,dt}}=\left[ {-{{t}^{{-1}}}} \right]_{2}^{{{{x}^{4}}}}=-{{\left( {{{x}^{4}}} \right)}^{{-1}}}-\left( {-{{2}^{{-1}}}} \right)=-\frac{1}{{{{x}^{4}}}}+\frac{1}{2}$$ Now take the derivative back: $$\displaystyle \frac{d}{{dx}}\left( {-\frac{1}{{{{x}^{4}}}}+\frac{1}{2}} \right)=\frac{d}{{dx}}\left( {-{{x}^{{-4}}}} \right)+\frac{d}{{dx}}\left( {\frac{1}{2}} \right)=4{{x}^{{-5}}}+0=\frac{4}{{{{x}^{5}}}}$$   √ Find the derivative of $$\displaystyle F\left( x \right)=\int\limits_{{-\pi }}^{{{{x}^{2}}}}{{\sin {{t}^{2}}\,dt}}$$ $$\displaystyle {F}’\left( x \right)=\sin {{\left( {{{x}^{2}}} \right)}^{2}}\cdot \frac{d}{{dx}}\left( {{{x}^{2}}} \right)=\sin \left( {{{x}^{4}}} \right)\cdot 2x=2x\sin {{x}^{4}}$$ Find the derivative of $$\displaystyle F\left( x \right)=\int\limits_{x}^{{3x}}{{2{{t}^{3}}\,dt}}$$ Since we don’t have a constant for the lower bound, we need to separate the integral into two integrals, each with a constant. Remember that we can switch the lower and upper bounds in an integral by taking the opposite (the negative) of that integral: $$\displaystyle \int\limits_{x}^{{3x}}{{2{{t}^{3}}\,dt}}=\int\limits_{x}^{a}{{2{{t}^{3}}\,dt}}+\int\limits_{a}^{{3x}}{{2{{t}^{3}}\,dt}}=-\int\limits_{a}^{x}{{2{{t}^{3}}\,dt}}+\int\limits_{a}^{{3x}}{{2{{t}^{3}}\,dt}}$$ \begin{align}{F}’\left( x \right)&=-2{{x}^{3}}\cdot \frac{d}{{dx}}\left( x \right)+2{{\left( {3x} \right)}^{3}}\cdot \frac{d}{{dx}}\left( {3x} \right)\\&=-2{{x}^{3}}\cdot 1+2\left( {27{{x}^{3}}} \right)\cdot 3=-2{{x}^{3}}+162{{x}^{3}}=160{{x}^{3}}\end{align}

# Integration as Accumulated Change

As the result of the Fundamental Theorems of Calculus, we can now use integration to solve Accumulated Rate of Change, or Net Change problems.

Here is the Net Change Theorem (which is basically a reinstatement of the Fundamental Theorem of Calculus):

The definite integral of the rate of change gives the total accumulated change, or net change, of the quantity in interval $$\left[ {a,b} \right]$$:

$$\displaystyle \int\limits_{a}^{b}{{{F}’\left( x \right)}}\,dt=F\left( b \right)-F\left( a \right)$$

The Accumulated Rate of Change can be measured by the area under the graph of a function over a certain interval. We must be careful though, since the area below the $$\boldsymbol {x}$$-axis is considered to be negative in measuring accumulated change. Thus, this can be represented by the definite integral of the function.

Note that we also addressed Position, Velocity, and Acceleration with Derivatives here in the Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change section, and here in the Antiderivatives and Indefinite Integration section.

Important Hint for Definite Integration Applications: If you’re not sure about whether to integrate, or what to integrate, remember this: the area under the curve is the integral. That area will represent different things, based on what the units of the axes are. In general, you just need to multiply the units of the $$y$$-axis and the units of the $$x$$-axis to get the units of the area under the curve. For example, If the $$y$$-axis represents velocity and the $$x$$-axis represents time, the integral represents total distance. As another example, if the $$y$$-axis represents calories/day and the $$x$$-axis represents days, then the area under the curve would represent the total calories over the days specified.

When working these problems, remember the following:

Integration as Accumulated Change Hints

• For Integration as Accumulated Change problems, we typically have rate (velocity) on the $$y$$-axis and time on the $$x$$-axis. The change is the area under the curve, or the integral of the velocity function. For example, we may have:

$$\require{cancel} \displaystyle \frac{{\text{miles}}}{{\cancel{{\text{hours}}}}}\text{(}y\text{-axis)}\times \cancel{{\text{hours}}}\text{(}x\text{-axis)}=\text{miles}$$

• Amount changed from $$a$$ to $$b$$ is $$\displaystyle \int\limits_{a}^{b}{{{f}’\left( x \right)}}\,dx$$, where $${f}’\left( x \right)$$ is the rate of change of the amounts. $$f\left( x \right)$$ represents the actual amounts (for example, miles, dollars or temperature).
• Ending amount is the beginning amount plus the change. At $$b$$, the amount is what it was at $$a$$ (Initial Condition), plus the change from $$a$$ to $$b$$:  $$\displaystyle f\left( b \right)=f\left( a \right)+\int\limits_{a}^{b}{{{f}’\left( x \right)}}\,dx$$  (from First Fundamental Theorem of Calculus). If change is negative (amount decreasing), $$\displaystyle f\left( b \right)=f\left( a \right)-\int\limits_{a}^{b}{{{f}’\left( x \right)}}\,dx$$.
• The Average Velocity can be obtained by our “Integral Over Interval” formula: Average Velocity from time $$a$$ to time $$b$$ is $$\displaystyle \frac{{\int\limits_{a}^{b}{{f\left( x \right)\,dx}}}}{{\left( {b-a} \right)}}=\frac{1}{{\left( {b-a} \right)}}\int\limits_{a}^{b}{{f\left( x \right)\,dx}}$$, where $$f\left( x \right)$$ is a function of the velocity versus time.
• The total distance traveled (how far we go back and forth) is $$\displaystyle \int\limits_{a}^{b}{{\left| {\,v\left( x \right)} \right|}}\,dx$$, whereas the total displacement (where we are on a line, compared to where we started) is $$\displaystyle \int\limits_{a}^{b}{{v\,\left( x \right)}}\,dx$$.
• The Definite Integral of a function’s population growth rate in a time interval gives the total change in population in that interval. The Definite Integral of a density of a population in a distance interval gives the total amount of that population in that interval.

Note that if the function is totally above the $$y$$-axis in the given interval, this calculation is the area between the function and the $$x$$-axis. If the function is totally below the $$x$$-axis in this interval, this calculation is the opposite (negative) of the area between the $$x$$-axis and the function. If this function is both above and below the $$x$$-axis, then this calculation is the area above the $$x$$-axis subtracted by the area below the $$x$$-axis.

Remember these rules with position, velocity, and acceleration:

• Definite Integral of a function’s derivative gives the accumulated change.
• Definite Integral of a function’s velocity gives the total change in position.
• Definite Integral of a function’s acceleration gives the total change in velocity.
• When the velocity is positive, an article is moving to the right, when it’s negative, it’s moving to the left, and when it’s 0, it’s standing still. This makes sense since the area under the curve of a velocity graph is distance, and when distance is positive, it’s accumulating (article is moving to right), and when it’s negative, it’s lessening (article moving to left).

First let’s do a problem just by looking just the area under the curve:

Problem:

The following graph depicts the speed of a car over a period over 6 hours. Estimate how far it traveled during that time, and the average speed between 10am and 4pm. Solution:

Since we can think of distance as $$\text{rate}\,\times \,\text{time}$$, we can get the distance by the area under the curve. Add up (as best we can) all the squares under the curve: $$\left( {20\times 12} \right)+18+14+8+8+9+8+5+8+16=334\,\,\text{km}$$.

To get the average speed between 10am and 4pm,take the total distance and divide by the total time: $$\displaystyle \frac{{334\,\text{km}}}{{6\,\text{hours}}}=55.7\,\text{km/hour}$$.

## More Accumulated Change Problems

Here’s another type of problem you may see:

Accumulating Change Problem:

A particle is moving along the $$x$$-axis, with the particle’s velocity (units/sec) for time $$t$$ ($$0\le t\le 9$$) as shown: a. Find the total displacement of the particle in the 9-second interval.

b. Find the particle’s total distance traveled in the 9-second interval.

c. If the position of the particle is $$p\left( t \right)$$, and $$p\left( 4 \right)=3$$, find the position of the particle at $$t=9$$.

d. Find the average value of the velocity of the particle during this 9-second time interval.

e. What is the displacement of the particle from the 2nd second to the 6th second? (from $$t=1$$ to $$t=6$$)?

Solution:

a. The displacement is the value of the integral of the function in that interval, with the areas above the $$x$$-axis positive and under the $$x$$-axis negative. This displacement is $$8-8=0$$.

b. The distance is the absolute value of the integral across this interval. The distance traveled is $$8+8=16$$.

c. If the position at $$t=4$$ is 3, the position at $$t=9$$ would be $$\displaystyle 3+\frac{1}{2}-8=-4\frac{1}{2}$$. This is because the ending value is the beginning value (3) plus the change, which is the integral. We start out at 3, then have above the $$x$$-axis (from $$t=4$$ to $$t=5$$) and 8 below the $$x$$-axis (from $$t=5$$ to $$t=9$$).

d. The average value is integral (displacement) over interval, so we have $$\displaystyle \frac{0}{9}=0$$ (which is the same as $$\displaystyle \frac{1}{{\left( {b-a} \right)}}\int\limits_{a}^{b}{{f\left( t \right)\,dt}}=\frac{1}{{\left( {9-0} \right)}}\int\limits_{0}^{9}{{f\left( t \right)\,dt}}=\frac{0}{9}=0$$).

e. The displacement from the 2nd second to the 6th second is the area (positive or negative) from where $$t=1$$ to $$t=6$$, which is $$6 -.5=5.5$$. This is the same as $$\displaystyle \int\limits_{1}^{6}{{f\left( t \right)\,dt}}=-5.5$$. Note that we started the interval with $$t=1$$ instead of $$t=2$$, since the 2nd second starts when $$t=1$$.

Here are more Integration as Accumulated Change Problems. Note that in some problems, we will use the fnInt( function on our calculator to integrate. (We will learn how to integrate exponents here in the Exponential and Logarithmic Integration section.)

 Integration as Accumulated Change Problem Solution In 2008, the average value of an American’s income could be changing at a rate (dollars per month) by the function $$c\left( t \right)=40{{\left( {1.002} \right)}^{t}}$$, where $$t$$ is months since January 1, 2008.    What change in income can the average American expect by the end of September, 2008? Since we want the change in income, we’ll use the formula $$\int\limits_{a}^{b}{{{f}’\left( t \right)}}\,dt$$, where $$f\left( t \right)$$ is the actual amount of income.   We have the rate of change formula $$c\left( t \right)=40{{\left( {1.002} \right)}^{t}}$$, so we’ll integrate that function from 0 (January) to 9 (September) (use calculator): $$\int\limits_{0}^{9}{{{f}’\left( t \right)}}\,dt=\int\limits_{0}^{9}{{40{{{\left( {1.002} \right)}}^{t}}}}\,dt=\363.26$$. Water is leaking out of a bucket at a rate of $$r\left( t \right)=6{{e}^{{-.1t}}}$$ liters per minute, where $$t$$ is the number of minutes since the leak started.   If the bucket holds 500 liters of water when the leak began, how much water does the bucket hold an hour later? Ending amount is the beginning amount (initial condition) plus the change. At $$b$$, the amount is what it was at $$a$$, plus the change from $$a$$ to $$b$$. Since water is leaking out of the bucket (negative change), its amount is decreasing: $$f\left( b \right)=f\left( a \right)-\int\limits_{a}^{b}{{{f}’\left( x \right)}}\,dx$$. So $$f\left( b \right)=500-\int\limits_{0}^{{60}}{{6{{e}^{{-.1t}}}}}\,dt=500-59.851=440.149$$ liters. (Use calculator to integrate and remember that 60 minutes are in an hour.) The density of cars (cars per mile) on a 10-mile stretch can be modeled by $$d\left( x \right)=200\left[ {3+\sin \left( {4\sqrt{{x+1}}} \right)} \right]\,$$, where $$x$$ is the distance in miles from the starting point.   To the nearest car, what is the total number of cars on the 10-mile stretch? Remember that the total amount of a certain population in an interval can be obtained by integrating the density of that amount in that integral (see how $$\displaystyle \require{cancel} \frac{{\text{cars}}}{{\cancel{{\text{mile}}}}}\,\,\times \cancel{{\text{miles}}}=\text{cars}$$)? The total number of cars then on the 10-mile stretch is:                               $$\int\limits_{0}^{{10}}{{200\left[ {3+\sin \left( {4\sqrt{{x+1}}} \right)} \right]\,dx}}\approx 5716$$ cars. The velocity in meters per second of a particle is moving along the $$x$$-axis at $$v\left( t \right)=2\cos \left( t \right)\,$$ in the interval $$0\le t\le 2\pi$$.   Determine when the particle is moving to the left, right, and stopped.   Find the particle’s displacement and total distance over the time interval. When the velocity is positive, an article is moving to the right, when it’s negative, it’s moving to the left, and when it’s 0, it’s standing still. Thus, when the article is moving to the right, we have $$2\cos \left( t \right)\,>0$$, or $$\displaystyle 0 # Using U-Substitution with Definite Integration We learned how to use U-substitution here in the U-Substitution Integration section, but let’s revisit U-Sub and do some problems using Definite Integration. Remember this about U-sub Indefinite Integration: U-Substitution Integration Outside Function Derivative of Inside Function \(\displaystyle \int{{\color{red}{{f\left( {\color{green}{{g\left( x \right)}}} \right)}}\,\color{purple}{{{g}’\left( x \right)}}}}\,dx=\color{blue}{{F\left( {g\left( x \right)} \right)}}\,+\,C$$

Inside Function              Any Composite Function

What this says is that if we want the integral of the outside function, to make it work, we have to make sure that what we’re integrating somehow has a factor that is the derivative of the inside function. (We can “trick” the integrand into having this factor.)

With U-sub and Definite Integration, we can do these problems in one of two ways:

• We can go ahead and substitute the expression for “$$u$$” back and use the original values for the upper and lower bounds. Remember that the upper and lower bounds are in terms of “$$x$$”.
• We can keep the “$$u$$” in the expression and solve for new upper and lower bounds (solve for “$$x$$” in terms of “$$u$$”). Then we don’t have to put the expression for “$$u$$” back in the problem! (We only want to do this if it’s straightforward to get “$$x$$” in terms of “$$u$$”.)

We’ll show both these methods; the main thing is to make sure your upper and lower bounds match the variable you’re plugging them in for:

 Definite Integral U-Substitution Problem Solution Find the integral: $$\displaystyle \int\limits_{{-2}}^{2}{{x{{{\left( {{{x}^{2}}-3} \right)}}^{2}}\,dx}}$$ \displaystyle \begin{align}\color{blue}{{u={{x}^{2}}-3}}\\du=2x\,dx\\\color{green}{{dx=\frac{{du}}{{2x}}}}\end{align}          $$\require {cancel} \displaystyle \int_{{-2}}^{2}{{x{{{\left( {\color{blue}{{{{x}^{2}}-3}}} \right)}}^{2}}}}\color{green}{{dx}}=\int_{{-2}}^{2}{{\cancel{x}\cdot {{{\color{blue}{u}}}^{2}}}}\color{green}{{\frac{{du}}{{2\cancel{x}}}}}=\int_{{-2}}^{2}{{\frac{{{{{\color{blue}{u}}}^{2}}}}{2}\color{green}{{du}}}}=\left[ {\frac{{{{u}^{3}}}}{6}} \right]_{{x=-2}}^{{x=2}}$$ We’ll substitute back in for “$$u$$” since it’s not easy to solve for $$x$$ in terms of $$u$$: $$\displaystyle \left[ {\frac{{{{u}^{3}}}}{6}} \right]_{{x=-2}}^{{x=2}}=\left[ {\frac{{{{{\left( {{{x}^{2}}-3} \right)}}^{3}}}}{6}} \right]_{{-2}}^{2}=\frac{{{{{\left( {{{2}^{2}}-3} \right)}}^{3}}}}{6}-\frac{{{{{\left( {{{{\left( {-2} \right)}}^{2}}-3} \right)}}^{3}}}}{6}=\frac{1}{6}-\frac{1}{6}=0$$ Find the integral:   $$\displaystyle \int\limits_{{-\frac{\pi }{2}}}^{{\frac{\pi }{2}}}{{{{{\sin }}^{3}}\left( x \right)\cos \left( x \right)\,dx}}$$ $$\displaystyle \begin{array}{l}\\\color{blue}{{u=\sin \left( x \right)}}\\du=\cos \left( x \right)\,dx\\\color{green}{{dx=\frac{{du}}{{\cos \left( x \right)}}}}\end{array}$$           $$\displaystyle \int\limits_{{-\frac{\pi }{2}}}^{{\frac{\pi }{2}}}{{{{{\color{blue}{{\sin }}}}^{3}}\color{blue}{{\left( x \right)}}\cos \left( x \right)\,\color{green}{{dx}}}}=\int\limits_{{-\frac{\pi }{2}}}^{{\frac{\pi }{2}}}{{{{{\color{blue}{u}}}^{3}}\,\cancel{{\cos \left( x \right)}}\,\color{green}{{\frac{{du}}{{\cancel{{\cos \left( x \right)}}}}}}}}=\int\limits_{{-\frac{\pi }{2}}}^{{\frac{\pi }{2}}}{{{{{\color{blue}{u}}}^{3}}\color{green}{{du}}}}$$ $$\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left[ {\frac{{{{u}^{4}}}}{4}} \right]_{{x=-\frac{\pi }{2}}}^{{x=\frac{\pi }{2}}}=\left[ {\frac{{{{{\sin }}^{4}}\left( x \right)}}{4}} \right]_{{-\frac{\pi }{2}}}^{{\frac{\pi }{2}}}=\frac{{{{{\sin }}^{4}}\left( {\frac{\pi }{2}} \right)}}{4}-\frac{{{{{\sin }}^{4}}\left( {-\frac{\pi }{2}} \right)}}{4}=0$$ Find the integral:   $$\displaystyle \int\limits_{1}^{4}{{\frac{2}{{\sqrt{x}{{{\left( {1+\sqrt{x}} \right)}}^{2}}}}}}$$ $$\displaystyle \begin{array}{l}\color{blue}{{u=1+\sqrt{x}=1+{{x}^{{\frac{1}{2}}}}}}\\du=\frac{1}{2}{{x}^{{-\frac{1}{2}}}}dx\\\color{green}{{dx=\,\frac{{du}}{{\frac{1}{2}{{x}^{{-\frac{1}{2}}}}}}=2{{x}^{{\frac{1}{2}}}}\,du}}\end{array}$$ $$\displaystyle \int\limits_{1}^{4}{{\frac{2}{{\sqrt{x}{{{\left( {\color{blue}{{1+\sqrt{x}}}} \right)}}^{2}}}}}}\,\color{green}{{dx}}=\int\limits_{1}^{4}{{\frac{2}{{\sqrt{x}\,\color{blue}{{{{u}^{2}}}}}}}}\,\color{green}{{dx}}=\int\limits_{1}^{4}{{\frac{2}{{\cancel{{{{x}^{{\frac{1}{2}}}}}}\,\color{blue}{{{{u}^{2}}}}}}}}\color{green}{{\,\left( {2\cancel{{{{x}^{{\frac{1}{2}}}}}}\,du} \right)}}=\int\limits_{1}^{4}{{4\color{blue}{{{{u}^{{-2}}}}}\color{green}{{du}}}}\,$$ $$\displaystyle =\left[ {\frac{{4{{u}^{{-1}}}}}{{-1}}} \right]_{{x=1}}^{{x=4}}=\left[ {\frac{{-4}}{u}} \right]_{{x=1}}^{{x=4}}=\left[ {\frac{{-4}}{{1+\sqrt{x}}}} \right]_{1}^{4}=\left( {\frac{{-4}}{{1+\sqrt{4}}}} \right)-\left( {\frac{{-4}}{{1+\sqrt{1}}}} \right)=\frac{2}{3}$$ Find the integral:   $$\displaystyle \int\limits_{{-1}}^{2}{{x\sqrt{{x+2}}\,dx}}$$   For this problem, find the integral by replacing the upper and lower bounds with value of “$$u$$”. $$\displaystyle \begin{array}{l}\color{blue}{{u=x+2}}\\\color{red}{{x=u-2}}\\du=dx\\\color{green}{{dx=du}}\end{array}$$ $$\displaystyle \int_{{-1}}^{2}{{\cancel{x}\sqrt{{\color{blue}{{x+2}}}}\,\color{green}{{dx}}=\int_{{-1}}^{2}{{\color{red}{x}{{{\left( {\color{blue}{{x+2}}} \right)}}^{{\frac{1}{2}}}}\,\color{green}{{dx}}=}}}}\int_{{-1}}^{2}{{\color{red}{{\left( {u-2} \right)}}\cdot {{{\color{blue}{u}}}^{{\frac{1}{2}}}}}}\color{green}{{du}}=\int_{{-1}}^{2}{{\left( {\color{blue}{{u\cdot {{u}^{{\frac{1}{2}}}}-2{{u}^{{\frac{1}{2}}}}}}} \right)}}\,\color{green}{{du}}$$ $$\displaystyle =\int_{{-1}}^{2}{{\left( {\color{blue}{{{{u}^{{\frac{3}{2}}}}-2{{u}^{{\frac{1}{2}}}}}}} \right)}}\,\color{green}{{du}}=\left[ {\frac{{{{u}^{{\frac{5}{2}}}}}}{{\frac{5}{2}}}-2\frac{{{{u}^{{\frac{3}{2}}}}}}{{\frac{3}{2}}}} \right]_{{u=x+2=-1+2=1}}^{{u=x+2=2+2=4}}=\left[ {\frac{{2{{u}^{{\frac{5}{2}}}}}}{5}-\frac{{4{{u}^{{\frac{3}{2}}}}}}{3}} \right]_{1}^{4}$$ $$\displaystyle =\left( {\frac{{2{{{\left( 4 \right)}}^{{\frac{5}{2}}}}}}{5}-\frac{{4{{{\left( 4 \right)}}^{{\frac{3}{2}}}}}}{3}} \right)-\left( {\frac{{2{{{\left( 1 \right)}}^{{\frac{5}{2}}}}}}{5}-\frac{{4{{{\left( 1 \right)}}^{{\frac{3}{2}}}}}}{3}} \right)=\frac{{46}}{{15}}$$

Understand these problems, and practice, practice, practice!

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On to Exponential and Logarithmic Integration – you’re ready!

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