# Integral Calculus Quick Study Guide

Note: The Differential Calculus Quick Study Guide can be found here.

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 CALCULUS Quick Study Guide: INTEGRATION Basic Integrals: $$\displaystyle \begin{array}{l}\int{0}\,\,dx=\text{C}\\\int{k}\,dx=kx+C\end{array}$$ $$\displaystyle \int{{{{x}^{n}}}}dx=\frac{{{{x}^{{n+1}}}}}{{n+1}}\,+C,\,\,n\ne 1$$ $$\displaystyle \begin{array}{l}\int{{k\cdot f\left( x \right)dx}}=k\int{{f\left( x \right)dx}}\\\int{{\left[ {f\left( x \right)\,\,\pm \,\,g\left( x \right)} \right]}}\,\,dx\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int{{f\left( x \right)dx\pm \,\,\int{{g\left( x \right)}}}}\,dx\end{array}$$   Trig Integrals: $$\begin{array}{l}\int{{\cos \,x}}\,dx=\sin x+C\\\int{{\sin \,x\,}}dx=-\cos x+C\\\int{{{{{\sec }}^{2}}\,x}}\,dx=\tan x+C\\\int{{\sec \,x\,\,\tan x}}\,dx=\sec x+C\\\int{{{{{\csc }}^{2}}x}}\,dx=-\cot x+C\\\int{{\csc x\,\,\cot x}}\,dx=-\csc x+C\end{array}$$   (Trig functions starting with “c” are negative. I also remember that there are always 2 sec’s (csc’s) and 1 tan (cot’s) in the last 4 equations.)   Integrals of the Six Basic Trig Functions: $$\begin{array}{l}\int{{\sin u\,du=-\cos u\,\,+\,\,C}}\\\int{{\tan u\,du=-\ln \left| {\cos u} \right|\,\,+\,\,C}}\\\int{{\sec u\,du=\ln \left| {\sec u+\tan u} \right|\,\,+\,\,C}}\\\int{{\cos u\,du=\sin u\,\,+\,\,C}}\\\int{{\cot u\,du=\ln \left| {\sin u} \right|\,\,+\,\,C}}\\\int{{\csc u\,du=-\ln \left| {\csc u+\cot u} \right|\,\,+\,\,C}}\end{array}$$ Integrals Involving Inverse Trig Functions: \begin{align}&\int{{\frac{{du}}{{\sqrt{{{{a}^{2}}-{{u}^{2}}}}}}}}\,=\arcsin \,\frac{u}{a}+C\\&\int{{\frac{{du}}{{{{a}^{2}}+{{u}^{2}}}}}}\,=\frac{1}{a}\arctan \,\frac{u}{a}+C\\&\int{{\frac{{du}}{{u\sqrt{{{{u}^{2}}-{{a}^{2}}}}}}}}\,=\frac{1}{a}\text{arcsec}\,\frac{{\left| u \right|}}{a}+C\end{align}   U-sub Integration: $$\int{{f\left( {g\left( x \right)} \right){g}’\left( x \right)\,\,}}dx=F\left( {g\left( x \right)} \right)\,+\,\,C$$ 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.)   U-sub Simplification Formula: $$\displaystyle \int{{{{{\left( {ax+b} \right)}}^{n}}\,dx=\frac{{{{{\left( {ax+b} \right)}}^{{n+1}}}}}{{a\left( {n+1} \right)}}}}+\,C$$   Fundamental Theorem of Calculus: For $$\displaystyle {F}’\left( x \right)=f\left( x \right),\,\,\,\,\int\limits_{a}^{b}{{f\left( x \right)dx}}=F\left( a \right)-F\left( b \right)$$   2nd Fundamental Theorem of Calculus: If a is a constant, and $$\displaystyle F\left( x \right)=\int\limits_{a}^{x}{{f\left( t \right)\,dt}}$$, then $${F}’\left( x \right)=f\left( x \right)$$. (HINT: To find $${F}’\left( x \right)$$, 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 what you get by the derivative of the upper bound.) L’Hôpital’s Rule: If f and g are differentiable functions and if $$\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)}}{{g\left( x \right)}}\,\,\text{is}\,\,\frac{0}{0}\,\,\text{or}\,\,\frac{\infty }{\infty }\,\,\text{or}\,\,\frac{{-\infty }}{{-\infty }}\,\,\text{or}\,\,\frac{{-\infty }}{\infty }\,\,\text{or}\,\,\frac{\infty }{{-\infty }}$$, then  $$\displaystyle \underset{{x\to c}}{\mathop{{\lim }}}\,\,\frac{{f\left( x \right)}}{{g\left( x \right)}}\,\,=\,\,\underset{{x\to c}}{\mathop{{\lim }}}\,\,\frac{{{f}’\left( x \right)}}{{{g}’\left( x \right)}}=\,\,\underset{{x\to c}}{\mathop{{\lim }}}\,\,\frac{{{f}^{\prime \prime}(x)}}{{{g}^{\prime \prime}(x)}},\,\,\text{etc}$$                  (as long as the limit on the right exists or is infinite). What this says is that if you have two functions whose limits fit the conditions above, you can take the derivative of the top function and bottom function (separately) over and over again until you get limits that “work”.   Definition of the Area of a Region by a Limit: Let f  be continuous and above the y-axis (non-negative) on interval $$\left[ {a,\,b} \right]$$. The area of the region bounded by f, the x-axis and vertical lines at $$x=a$$ and $$x=b$$ is: \displaystyle \begin{align}\text{Area}&=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{f\left( {{{c}_{i}}} \right)\cdot \Delta x}}\text{, where }\Delta x=\frac{{b-a}}{n},\text{ and}\,\text{ }{{x}_{{i-1}}}\le {{c}_{i}}\le {{x}_{i}}\\&=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{f\left( {a+\frac{{\left( {b-a} \right)i}}{n}} \right)\cdot \left( {\frac{{b-a}}{n}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(=\int\limits_{a}^{b}{{f\left( x \right)}}\,dx,\,\text{when above }y\text{ axis})\\\,\,\,&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{height}\,\,\,\,\,\,\,\text{x}\,\,\,\,\,\,\text{width}\end{align} The Trapezoidal Rule: Let f  be continuous on interval $$\left[ {a,\,b} \right]$$. We can approximate $$\int\limits_{a}^{b}{{f\left( x \right)}}\,dx$$ by the Trapezoidal Rule: \displaystyle \begin{align}\int\limits_{a}^{b}{{f\left( x \right)}}\,dx\,\,\approx \,\,\frac{{b-a}}{{2n}}\left[ {f\left( {{{x}_{0}}} \right)+2f\left( {{{x}_{1}}} \right)+…+2f\left( {{{x}_{{n-1}}}} \right)+f\left( {{{x}_{n}}} \right)} \right],\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{where }f\left( {{{x}_{0}}} \right)=f\left( a \right),\,\,\,f\left( {{{x}_{1}}} \right)=f\left( {a+\frac{{b-a}}{n}} \right),\,\,\text{and so on}\end{align} Mean Value Theorem for Integrals: For a function f  that is continuous on closed interval $$\left[ {a,\,b} \right]$$, there exists at least one number c in that closed interval such that  $$\int\limits_{a}^{b}{{f\left( x \right)\,dx=f\left( c \right)}}\left( {b-a} \right)$$. (You may have to find the “c” by setting $$f\left( c \right)$$ to $$\displaystyle \frac{{\int\limits_{a}^{b}{{f\left( x \right)\,dx}}}}{{\left( {b-a} \right)}}$$, solving for c, eliminating any values not in interval $$\left[ {a,\,b} \right]$$.   Average Value of a Function: If a function f  is integrable on a closed interval $$\left[ {a,\,b} \right]$$, 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:   Average Value = .$$\displaystyle \frac{{\int\limits_{a}^{b}{{f\left( x \right)\,dx}}}}{{b-a}}=\,\,\frac{{\text{Integral}}}{{\text{Interval}}}$$.   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 $$\require{cancel} \displaystyle \frac{{\text{miles}}}{{\cancel{{\text{hours}}}}}\text{ (}y\text{-axis)}\,\,\times \,\,\cancel{{\text{hours}}}\text{(}x\text{-axis)}\,\,=\text{ miles}$$. The Average Velocity may be obtained by the “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 $$\int\limits_{a}^{b}{{\left| {\,v\left( x \right)} \right|}}\,dx$$, 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 gives the total change in population. ·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. Definite Integral as the Area of a Region: Let f  be continuous and above the y-axis (non-negative) on interval $$\left[ {a,\,b} \right]$$. 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:   $$\displaystyle \text{Area}=\int\limits_{a}^{b}{{f\left( x \right)}}\,dx$$. Note:  The area of the region represented by an integral is only applicable if the region in the interval is totally above the x-axis (positive y). Any region below the x-axis represents a “negative area”.   Exponential Growth and Decay Formula: For a function $$y>0$$ that is differentiable function of t, and $${y}’=ky$$, $$y=C{{e}^{{kt}}}$$ C is the initial value of y, k is the proportionality constant. For $$k>0$$, we have exponential growth, and for $$k<0$$, we have exponential decay.   Area of Region Between Two Curves: For functions f and g where $$f\left( x \right)\ge g\left( x \right)$$ for all x in $$\left[ {a,\,b} \right]$$, the area of the region bounded by the graphs and the vertical lines x = a and x = b is:  $$\text{Area}=\int\limits_{a}^{b}{{\left[ {f\left( x \right)-g\left( x \right)} \right]}}\,dx$$.   Volumes of Solids: The Disk Method: Horizontal Axis:    $$\text{Volume}=\pi \,\int\limits_{a}^{b}{{{{{\left[ {f\left( x \right)} \right]}}^{2}}}}\,dx$$            Vertical Axis:    $$\text{Volume}=\pi \,\int\limits_{a}^{b}{{{{{\left[ {f\left( y \right)} \right]}}^{2}}}}\,dy$$   Volumes of Solids: The Washer Method:   Horizontal Axis:   $$\text{Volume}=\pi \int\limits_{a}^{b}{{\left( {{{{\left[ {R\left( x \right)} \right]}}^{2}}-{{{\left[ {r\left( x \right)} \right]}}^{2}}} \right)}}\,dx$$            Vertical Axis:   $$\text{Volume}=\pi \,\int\limits_{a}^{b}{{\left( {{{{\left[ {R\left( y \right)} \right]}}^{2}}-{{{\left[ {r\left( y \right)} \right]}}^{2}}} \right)}}\,\,dy$$   Volumes of Solids: The Shell Method: Revolution around the y-axis:     $$\text{Volume}=2\pi \int\limits_{a}^{b}{{x\,f\left( x \right)}}\,dx$$        Revolution around the x-axis:  $$\text{Volume}=2\pi \int\limits_{a}^{b}{{y\,f\left( y \right)}}\,dy$$ Integration by Parts: For functions u and v that are functions of x with continuous derivatives, then: $$\displaystyle \int{{u\,dv=uv\,\,-}}\,\int{{v\,}}du$$ Remember LIATE when trying to choose what to pick for u (in this order):  Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, and Exponential Functions
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