This section covers:
 Introduction to Definite Integrals
 Properties of Definite Integrals
 1st Fundamental Theorem of Calculus
 Definite Integrals on the Graphing Calculator
 Definite Integration and Area
 Mean Value Theorem (MVT) for Integrals
 Average Value of a Function
 Integration as Accumulated Change and Average Value Applications
 2nd Fundamental Theorem of Calculus
 Using USubstitution with Definite Integration
 More Practice
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 (nonnegative) 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:
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).
Problem  Solutions 
Evaluate the integrals, given the following values:
\(\int\limits_{3}^{5}{{dx}}=4\,\,\,\,\,\,\,\,\,\,\int\limits_{3}^{5}{{x\,dx}}=10\,\,\,\,\,\,\,\,\,\,\int\limits_{3}^{5}{{{{x}^{2}}dx}}=50\)
a. \(\int\limits_{5}^{3}{{x\,dx}}\) b. \(\int\limits_{3}^{5}{{4x\,dx}}\) c. \(\int\limits_{3}^{5}{{\left( {\frac{1}{2}{{x}^{2}}+3x8} \right)dx}}\) 
a. \(\int\limits_{5}^{3}{{x\,dx}}=\int\limits_{3}^{5}{{x\,dx}}=10\)
b. \(\int\limits_{3}^{5}{{4x\,dx}}=4\int\limits_{3}^{5}{{x\,dx}}=\left( 4 \right)\left( {10} \right)=40\) c. \(\begin{array}{l}\int\limits_{3}^{5}{{\left( {\frac{1}{2}{{x}^{2}}+3x8} \right)\,dx}}=\frac{1}{2}\int\limits_{3}^{5}{{{{x}^{2}}dx+3\int\limits_{3}^{5}{{x\,dx8\int\limits_{3}^{5}{d}}}x}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}\left( {50} \right)+3\left( {10} \right)8\left( 4 \right)=23\end{array}\) 
Evaluate the integrals, given the following values:
\(\int\limits_{1}^{4}{{f\left( x \right)dx}}=5\,\,\,\,\,\,\,\,\,\,\,\int\limits_{4}^{8}{{f\left( x \right)dx}}=3\,\,\,\,\,\,\,\,\,\,\,\int\limits_{4}^{8}{{g\left( x \right)dx}}=15\)
a. \(\int\limits_{1}^{8}{{f\left( x \right)\,dx}}\) b. \(\int\limits_{4}^{1}{{4f\left( x \right)\,dx}}\) c. \(\int\limits_{4}^{8}{{\left[ {3f\left( x \right)g\left( x \right)} \right]\,dx}}\) d. \(\int\limits_{4}^{4}{{4f\left( x \right)\,dx}}\) 
a. \(\int\limits_{1}^{8}{{f\left( x \right)\,dx}}=\int\limits_{1}^{4}{{f\left( x \right)\,dx}}+\int\limits_{4}^{8}{{f\left( x \right)\,dx}}=5+3=2\)
b. \(\begin{array}{l}\int\limits_{4}^{1}{{4f\left( x \right)\,dx}}=4\int\limits_{4}^{1}{{f\left( x \right)x\,dx}}=4\left( {\int\limits_{1}^{4}{{f\left( x \right)x\,dx}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( {4} \right)\left( {5} \right)=20\end{array}\) c. \(\displaystyle \begin{array}{l}\int\limits_{4}^{8}{{\left[ {3f\left( x \right)g\left( x \right)} \right]\,dx}}=3\int\limits_{4}^{8}{{f\left( x \right)\int\limits_{4}^{8}{{g\left( x \right)}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=3\left( {3} \right)15=24\end{array}\) d. \(\int\limits_{4}^{4}{{4f\left( x \right)\,dx}}=4\int\limits_{4}^{4}{{f\left( x \right)\,dx}}=4\left( 0 \right)=0\) 
1st Fundamental Theorem of Calculus
Wow! This sounds important, doesn’t it? That’s because the First 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:
First 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:
\(\displaystyle \int\limits_{a}^{b}{{f\left( x \right)}}\,dx=F\left( a \right)F\left( b \right)\)
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”:
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 
\(\displaystyle \int\limits_{1}^{3}{{5x\,dx}}\) 
\(\displaystyle \int\limits_{1}^{3}{{5x\,dx}}=\left[ {\frac{{5{{x}^{2}}}}{2}} \right]_{1}^{3}=\frac{{5{{{\left( 3 \right)}}^{2}}}}{2}\frac{{5{{{\left( 1 \right)}}^{2}}}}{2}=\frac{{45}}{2}\frac{5}{2}=20\) 
\(\displaystyle \int\limits_{0}^{8}{{\left( {\sqrt[3]{t}4} \right)\,dx}}\)  \(\displaystyle \begin{align}\int\limits_{0}^{8}{{\left( {{{t}^{{\frac{1}{3}}}}4} \right)\,dt}}&=\left[ {\frac{{{{t}^{{\frac{4}{3}}}}}}{{\frac{4}{3}}}4t} \right]_{0}^{8}=\left[ {\frac{{3{{t}^{{\frac{4}{3}}}}}}{4}4t} \right]_{0}^{8}\\&=\left( {\frac{{3{{{\left( 8 \right)}}^{{\frac{4}{3}}}}}}{4}4\left( 8 \right)} \right)\left( {\frac{{3{{{\left( 0 \right)}}^{{\frac{4}{3}}}}}}{4}4\left( 0 \right)} \right)=20\end{align}\) 
\(\displaystyle \int\limits_{0}^{{\frac{\pi }{2}}}{{\frac{{\cot x\cos x}}{{\cot x}}\,dx}}\)  \(\displaystyle \begin{align}\int\limits_{0}^{{\frac{\pi }{2}}}{{\frac{{\cot x\cos x}}{{\cot x}}\,dx}}&=\int\limits_{0}^{{\frac{\pi }{2}}}{{1\frac{{\cos x}}{{\left( {\frac{{\cos x}}{{\sin x}}} \right)}}\,=\int\limits_{0}^{{\frac{\pi }{2}}}{{1\sin x\,dx}}=\left[ {x+\cos x} \right]_{0}^{{\frac{\pi }{2}}}dx}}\\\,\,\,\,\,&=\left[ {\frac{\pi }{2}+\cos \left( {\frac{\pi }{2}} \right)} \right]\left[ {0+\cos \left( 0 \right)} \right]=\left( {\frac{\pi }{2}+0} \right)\left( {0+1} \right)\\&=\frac{\pi }{2}1\approx .571\,\,\,\,\,\end{align}\) 
\(\displaystyle \int\limits_{{5}}^{5}{{\left {{{x}^{2}}4} \right\,dx}}\) 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: \(\displaystyle \begin{array}{c}\int\limits_{{5}}^{5}{{\left {{{x}^{2}}4} \right\,dx}}=\int\limits_{{5}}^{{2}}{{\left( {{{x}^{2}}4} \right)\,dx+}}\int\limits_{{2}}^{2}{{\left( {{{x}^{2}}+4} \right)\,dx}}+\int\limits_{2}^{5}{{\left( {{{x}^{2}}4} \right)\,dx}}\\=\left[ {\left( {\frac{{{{x}^{3}}}}{3}4x} \right)} \right]_{{5}}^{{2}}+\left[ {\left( {\frac{{{{x}^{3}}}}{3}+4x} \right)} \right]_{{2}}^{2}+\left[ {\left( {\frac{{{{x}^{3}}}}{3}4x} \right)} \right]_{2}^{5}\\\,\,=\left[ {\left( {\frac{{{{{\left( {2} \right)}}^{3}}}}{3}4\left( {2} \right)} \right)\left( {\frac{{{{{\left( {5} \right)}}^{3}}}}{3}4\left( {5} \right)} \right)} \right]+\left[ {\left( {\frac{{{{{\left( 2 \right)}}^{3}}}}{3}+4\left( 2 \right)} \right)\left( {\frac{{{{{\left( {2} \right)}}^{3}}}}{3}+4\left( {2} \right)} \right)} \right]\\\,\,\,\,\,\,\,\,\,\,+\left[ {\left( {\frac{{{{{\left( 5 \right)}}^{3}}}}{3}4\left( 5 \right)} \right)\left( {\frac{{{{{\left( 2 \right)}}^{3}}}}{3}4\left( 2 \right)} \right)} \right]\\\,\,\,=27+\frac{{32}}{3}+27=\frac{{194}}{3}=64\frac{2}{3}\end{array}\) 
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.
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 xaxis 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:
Here are a few more problems on Definite Integration and Area:
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:
\(\displaystyle \int\limits_{a}^{b}{{f\left( x \right)\,dx=f\left( c \right)}}\left( {ba} \right)\).
For a geometric explanation, think of \(f\left( c \right)\) as the height of a rectangle and \(\left( {ba} \right)\) 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 \(\left( {ba} \right)\). So 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:
\(\displaystyle \frac{1}{{\left( {ba} \right)}}\int\limits_{a}^{b}{{f\left( x \right)\,dx}}\)
To remember this, think Integral over Interval:
\(\displaystyle \text{Average Value}=\frac{{\int\limits_{a}^{b}{{f\left( x \right)\,dx}}}}{{\left( {ba} \right)}}=\,\,\frac{{\text{Integral}}}{{\text{Interval}}}\)
Here are the types of problems you might see for the Mean Value Theorem:
Here are some Average Value Theorem problems:
Integration as Accumulated Change and Average Value Applications
One very useful application of Integration as Accumulated Change is that we can now use integration to solve Accumulated Rate of Change problems.
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.
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 \(a\), 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 \(\displaystyle \frac{{\int\limits_{a}^{b}{{f\left( x \right)\,dx}}}}{{\left( {ba} \right)}}=\frac{1}{{\left( {ba} \right)}}\int\limits_{a}^{b}{{f\left( x \right)\,dx}}\), 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.
And then 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.
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:
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 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 \(f\left( b \right)=500\int\limits_{0}^{{60}}{{6{{e}^{{.1t}}}}}\,dt=50059.851=440.149\) liters. (Use calculator to integrate and remember that 60 minutes are in an hour.) 
The density of cars (cars per mile) on a 10mile 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 10mile 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 10mile 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 \(\displaystyle 0<t<\frac{\pi }{2}\) and \(\displaystyle \frac{\pi }{2}<t<\frac{{3\pi }}{2}\). When the article is moving to the left, we have \(2\cos \left( t \right)\,<0\), or \(\displaystyle \frac{\pi }{2}<t<\frac{{3\pi }}{2}\). The particle is stopped at \(\displaystyle t=\frac{\pi }{2}\) and \(\displaystyle t=\frac{{3\pi }}{2}\). The particle’s displacement is \(\int\limits_{0}^{{2\pi }}{{2\cos \left( t \right)}}\,dt=0\) and the particle’s distance is \(\int\limits_{0}^{{2\pi }}{{\left {2\cos \left( t \right)} \right}}\,dt=8\). 
The rate at which customers arrive at a water park in the summer can be modeled by the function \(\displaystyle A\left( t \right)=\frac{{16000}}{{{{t}^{2}}22t+160}}\), and the rate at which they leave later that day can be modeled by the function \(\displaystyle L\left( t \right)=\frac{{10000}}{{{{t}^{2}}40t+350}}\) (both functions in people per hour, \(t\) is hours after midnight).
At \(t=10\) (right before the park opens), there are no people. At \(t=23\) (when park closes), everyone has left the park. (The park is open from 10am to 11pm.)


2^{nd} 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 \(\frac{d}{{dx}}\left( {\int\limits_{a}^{x}{{f\left( t \right)\,dt}}} \right)=f\left( x \right)\), 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: \(\frac{d}{{dx}}\left( {\int\limits_{a}^{{g\left( x \right)}}{{f\left( t \right)\,dt}}} \right)=\frac{d}{{dx}}\left( {F\left( {g\left( x \right)} \right)F\left( a \right)} \right)=\frac{d}{{dx}}F\left( {g\left( x \right)} \right)=f\left( {g\left( x \right)} \right)\cdot {g}’\left( x \right)\), 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 2^{nd} Fundamental Theorem of Calculus, with the basic instructions on how to solve these problems:
2^{nd} 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 \({F}’\left( x \right)=f\left( x \right)\)
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:
Using USubstitution with Definite Integration
We learned how to use Usubstitution here in the USubstitution Integration section, but let’s revisit USub and do some problems using Definite Integration.
Remember this about Usub Indefinite Integration:
USubstitution Integration
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 Usub 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:
Understand these problems, and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.
You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!
On to Exponential and Logarithmic Integration – you’re ready!