View source for Math 22 Integration by Substitution and the General Power Rule

==The General Power Rule==
  If <math>u</math> is a differentiable function of <math>x</math>, then
  <math>\int u^n\frac{du}{dx}dx=\int u^n du=\frac{u^{n+1}}{n+1}+C</math> for <math>n\ne -1</math>
==Guidelines for Integration by Substitution==
1.Let <math>u</math> be a function of <math>x</math> 

2.Rewrite the integral in terms of the variable <math>u</math>

3.Find the resulting integral in terms of <math>u</math>

4.Rewrite the antiderivative as a function of <math>x</math>

5.Check your answer by differentiating (optional)

'''Exercises 1''' Find the indefinite integral

'''1)''' <math>\int (5x^2+1)^2(10x)dx</math>
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution: &nbsp;
|-
|Let <math>u=5x^2+1</math>, so <math>du=10xdx</math>
|-
|<math>\int (5x^2+1)^2(10x)dx=\int u^2du=\frac{u^{2+1}}{2+1}+C=\frac{u^3}{3}+C=\frac{(5x^2+1)^3}{3}+C</math>
|}

'''2)''' <math>\int \sqrt{1-x^2}(-2x)dx</math>
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution: &nbsp;
|-
|Let <math>u=1-x^2</math>, so <math>du=-2xdx</math>
|-
|<math>\int \sqrt{1-x^2}(-2x)dx=\int \sqrt{u}du=\int u^{\frac{1}{2}}du=\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{2}{3}u^{\frac{3}{2}}+C=\frac{2}{3}(1-x^2)^{\frac{3}{2}}+C</math>
|}

'''3)''' Find an equation of the function <math>f(x)</math> that has the given derivative <math>f'(x)=2x(4x^2-10)^2</math> and whose graph passes through the point <math>(2,10)</math>
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution: &nbsp;
|-
|Let <math>u=4x^2-10</math>, so <math>du=8xdx</math>, so <math>dx=\frac{du}{8x}</math>
|-
|<math>f(x)=\int f'(x)dx=\int 2x(4x^2-10)^2dx=\int 2x(u)^2\frac{du}{8x}=\int\frac{1}{4}u^2du=\frac{1}{4}\int u^2du=\frac{1}{4}\frac{u^3}{3}+C=\frac{1}{12}(4x^2-10)^3+C</math>
|-
|Given <math>f(x)</math> passes through <math>(2,10)</math>, so <math>x=2,y=10</math> satisfy the equation <math>f(x)=\frac{1}{12}(4x^2-10)^3+C</math>.
|-
|So, <math>10=\frac{1}{12}(4(2)^2-10)^3+C</math>, hence <math>C=-8</math>
|-
|Therefore, <math>f(x)=\frac{1}{12}(4x^2-10)^3-8</math>.
|}



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