View source for Calculus/Indefinite integral/Solutions

{{question-answer|question=1. Evaluate <math>\int\frac{3x}{2}dx</math>|answer=We need to find a function, <math>F</math> , such that
:<math>F'(x)=\frac{3x}{2}</math>
We know that
:<math>\frac{d}{dx}x^2=2x</math>
So we need to find a constant, <math>a</math> , such that
:<math>\frac{d}{dx}ax^2=2ax=\frac{3x}{2}</math>
Solving for <math>a</math> , we get
:<math>2ax=\frac{3x}{2}\implies a=\frac{3}{4}</math>
So
:<math>\int\frac{3x}{2}=\frac{3}{4}x^2+C</math>
Check your answer by taking the derivative of the function you've found and checking that it matches the integrand:
:<math>\frac{d}{dx}\left(\frac{3}{4}x^{2}+C\right)=\frac{3}{4}(2x)=\frac{3x}{2}</math>}}

{{question-answer|question=2. Find the general antiderivative of the function <math>f(x)=2x^4</math>.|answer=We know that<br/>
<math>\frac{d}{dx}x^{5}=5x^{4}</math><br/>
We need to find a constant, <math>a</math>, such that<br/>
<math>\frac{d}{dx}ax^{5}=5ax^{4}=2x^{4}</math><br/>
Solving for <math>a</math>, we get<br/>
<math>5ax^{4}=2x^{4}\implies a=\frac{2}{5}</math><br/>
So the general antiderivative will be<br/>
<math>\mathbf{\frac{2}{5}x^{5}+C}</math><br/>
Check your answer by taking the derivative of the antiderivative you've found and checking that you get back the function you started with:<br/>
<math>\frac{d}{dx}\int 2x^4 dx=\frac{d}{dx}(\frac{2}{5}x^{5}+C)=\frac{2}{5}(5x^{4})=2x^{4}</math>}}

{{question-answer|question=3. Evaluate <math>\int(7x^2+3\cos(x)-e^x)dx</math>|answer=
<math>\begin{align}\int(7x^{2}+3\cos(x)-e^{x})dx&=7\int x^{2}dx+3\int\cos(x)dx-\int e^{x}dx\\
&=7(\frac{x^{3}}{3})+3\sin(x)-e^{x}+C\\
&=\mathbf{\frac{7}{3}x^3+3\sin(x)-e^x+C}\end{align}</math>}}

{{question-answer|question=4. Evaluate <math>\int(\frac{2}{5x}+\sin(x))dx</math>|answer=
<math>\begin{align}\int(\frac{2}{5x}+\sin(x))dx&=\frac{2}{5}\int\frac{dx}{x}+\int\sin(x)dx\\
&=\mathbf{\frac{2}{5}\ln|x|-\cos(x)+C}\end{align}</math>}}

{{question-answer|question=5. Evaluate <math>\int x\sin(2x^2)dx</math> by making the substitution <math>u=2x^2</math>|answer=Since <math>u=2x^{2}</math>, <math>du=4xdx</math> and <math>dx=\frac{du}{4x}</math><br/>
<math>\begin{align}\int x\sin(2x^{2})dx&=\int x\sin(u)\frac{du}{4x}\\
&=\frac{1}{4}\int\sin(u)du\\
&=-\frac{\cos(u)}{4}+C\\
&=-\mathbf{\frac{\cos(2x^{2})}{4}+C}\end{align}</math>}}

{{question-answer|question=6. Evaluate <math>\int-3\cos(x)e^{\sin(x)}dx</math>|answer=Let <math>u=\sin(x)</math>, <math>du=\cos(x)dx</math> so that <math>dx=\frac{du}{\cos(x)}</math><br/>
<math>\begin{align}\int-3\cos(x)e^{\sin(x)}dx&=-3\int\cos(x)e^{u}\frac{du}{\cos(x)}\\
&=-3\int e^{u}du\\
&=-3e^{u}+C\\
&=\mathbf{-3e^{\sin(x)}+C}\end{align}</math>}}

{{question-answer|question=7. Evaluate <math>\int \frac{2x-5}{x^3}dx</math> using integration by parts with <math>u=2x-5</math> and <math>dv=\frac{dx}{x^3}</math>|answer=<math>du=2dx</math>; <math>v=\int\frac{dx}{x^{3}}=-\frac{1}{2x^{2}}</math><br/>
<math>\begin{align}\int\frac{2x-5}{x^{3}}dx&=\int udv\\
&=uv-\int vdu\\
&=(2x-5)(-\frac{1}{2x^{2}})-\int(-\frac{1}{2x^{2}})2dx\\
&=\frac{5-2x}{2x^{2}}+\int\frac{dx}{x^{2}}\\
&=\frac{5-2x}{2x^{2}}-\frac{1}{x}\\
&=\frac{5-2x}{2x^{2}}-\frac{2x}{2x^{2}}\\
&=\mathbf{\frac{5-4x}{2x^{2}}}\end{align}</math>}}

{{question-answer|question=8. Evaluate <math>\int(2x-1)e^{-3x+1}dx</math>|answer=Let <math>u=2x-1</math>; <math>dv=e^{-3x+1}dx</math><br/>
Then <math>du=2dx</math> and <math>v=\int e^{-3x+1}dx</math><br/>
To evaluate <math>v</math>, make the substitution <math>w=-3x+1</math>; <math>dw=-3dx</math>; <math>dx=\frac{-dw}{3}</math>. Then<br/>
<math>v=\int e^{-3x+1}dx=\int e^{w}(\frac{-1}{3})dw=\frac{-e^{w}}{3}=\frac{-e^{-3x+1}}{3}</math>. So<br/>
<math>\begin{align}\int(2x-1)e^{-3x+1}dx&=\int udv\\
&=uv-\int vdu\\
&=(2x-1)\frac{-e^{-3x+1}}{3}-\int\frac{-e^{-3x+1}}{3}(2)dx\\
&=\frac{(1-2x)e^{-3x+1}}{3}+\frac{2}{3}\int e^{-3x+1}dx\\
&=\frac{(1-2x)e^{-3x+1}}{3}+\frac{2}{3}\int\frac{-e^{w}}{3}dw\\
&=\frac{3(1-2x)e^{-3x+1}}{9}-\frac{2}{9}e^{w}\\
&=\frac{(3-6x)e^{-3x+1}}{9}-\frac{2}{9}e^{-3x+1}\\
&=\mathbf{\frac{(1-6x)e^{-3x+1}}{9}}\end{align}</math> }}

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