Calculus/Indefinite integral/Solutions

From Wikibooks, open books for an open world
Jump to: navigation, search
1. Evaluate \int \frac{3x}{2}dx

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

2. Find the general antiderivative of the function f(x)=2x^4.

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

3. Evaluate \int(7x^2+3\cos(x)-e^x)dx

\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}

4. Evaluate \int(\frac{2}{5x}+\sin(x))dx

\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}

5. Evaluate \int x\sin(2x^2)dx by making the substitution u=2x^2

Since u=2x^{2}, du=4xdx and dx=\frac{du}{4x}
\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}

6. Evaluate \int-3\cos(x)e^{\sin(x)}dx

Let u=\sin(x), du=\cos(x)dx so that dx=\frac{du}{\cos(x)}
\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}

7. Evaluate \int \frac{2x-5}{x^3}dx using integration by parts with u=2x-5 and dv=\frac{dx}{x^3}

du=2dx; v=\int\frac{dx}{x^{3}}=-\frac{1}{2x^{2}}
\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}

8. Evaluate \int(2x-1)e^{-3x+1}dx

Let u=2x-1; dv=e^{-3x+1}dx
Then du=2dx and v=\int e^{-3x+1}dx
To evaluate v, make the substitution w=-3x+1; dw=-3dx; dx=\frac{-dw}{3}. Then
v=\int e^{-3x+1}dx=\int e^{w}(\frac{-1}{3})dw=\frac{-e^{w}}{3}=\frac{-e^{-3x+1}}{3}. So
\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}