Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 6

Integration

The Antiderivative and the General Power Rule

Observe that for every derivative we have an antiderivative.

1. Since ${\displaystyle {\displaystyle \frac{d}{dx}}{\left(x+1\right)}^2=2\left(x+1\right)\text{,}}$ we know that

   ${\displaystyle \int \left(x+1\right)dx={\displaystyle \frac{{\left(x+1\right)}^2}{2}}+C\text{.}}$

2. Similarly, since ${\displaystyle {\displaystyle \frac{d}{dx}}{\left(2x+1\right)}^3=6{\left(2x+1\right)}^2\text{,}}$ we know that

   ${\displaystyle \int {\left(2x+1\right)}^{2}dx={\displaystyle \frac{{\left(2x+1\right)}^3}{6}}+C\text{.}}$

3. Also, ${\displaystyle {\displaystyle \frac{d}{dx}}{\left(x^2+1\right)}^4=8x{\left(x^2+1\right)}^3\text{,}}$ so we know that

   ${\displaystyle \int x{\left(x^2+1\right)}^{3}dx={\displaystyle \frac{{\left(x^2+1\right)}^4}{8}}+C\text{.}}$

In all of these examples, we use the general power rule to differentiate; hence, the integrand function in all these cases is (or would be, except for a constant) a function of the form ${\displaystyle f{\left(x\right)}^r{f}^{\prime }\left(x\right)\text{.}}$

In (A), ${\displaystyle f\left(x\right)=\left(x+1\right)\text{,}}$ ${\displaystyle r=1}$ and ${\displaystyle f^{\prime }\left(x\right)=1\text{,}}$ so

${\displaystyle \int \left(x+1\right)dx={\displaystyle \int f(x)}}{f}^{\prime }(x)dx$.

In (B), ${\displaystyle f\left(x\right)=2x+1\text{,}}$ ${\displaystyle r=2\text{,}}$ and ${\displaystyle f^{\prime }\left(x\right)=2\text{,}}$ so

${\displaystyle \int {\left(2x+1\right)}^{2}dx={\displaystyle \frac{1}{2}}{\displaystyle \int {\left(2x+1\right)}^2\left(2\right)dx}}={\displaystyle \frac{1}{2}}{\displaystyle \int f{(x)}^r{f}^{\prime }(x)dx}$.

In (C), ${\displaystyle f\left(x\right)=\left(x^2+1\right)\text{,}}$ ${\displaystyle r=3}$ and $f^{\prime }\left(x\right)=2x$, so

${\displaystyle \int x{\left(x^2+1\right)}^{3}dx}={\displaystyle \frac{1}{2}}{\displaystyle \int {\left(x^2+1\right)}^3\left(2x\right)dx}={\displaystyle \frac{1}{2}}{\displaystyle \int f{\left(x\right)}^r{f}^{\prime }\left(x\right)dx}$.

From the general power rule, we know that

${\displaystyle \frac{d}{dx}}{\displaystyle \frac{f{\left(x\right)}^{r+1}}{r+1}}=f{\left(x\right)}^r{f}^{\prime }\left(x\right)$.

We conclude that

(6.6) ${\displaystyle \int f{(x)}^r{f}^{\prime }(x)dx={\displaystyle \frac{f{(x)}^{r+1}}{r+1}}+C}$   for   $r\ne -1$.

Note: An integral of the form $\color{#384877}{\displaystyle{\int \frac{f'(x)}{f(x)}\;dx}}$ is not covered in this course.

The antiderivatives presented above are found using Equation (6.6).

(A) \begin{align*} \int (x + 1)\; dx & = \int f(x)f'(x)\; dx\\ & = \dfrac{f(x)^2}{2} \\ & = \dfrac{(x + 1)^2}{2} + C. \end{align*}

(B) \begin{align*} \int (2x + 1)^2 \; dx & = \dfrac{1}{2}\int (2x + 1)^2 (2) \; dx \\ & = \dfrac{1}{2}\int f(x)^r f'(x) \; dx \\ & = \dfrac{1}{2}\dfrac{f(x)^3}{3} \\ & = \dfrac{(2x + 1)^3}{6} + C. \end{align*}

(C) \begin{align*} \int x(x^2 + 1)^3 \; dx & = \dfrac{1}{2} \int (x^2 + 1)^3 (2x) \;dx \\ & = \dfrac{1}{2} \int f(x)^r f'(x)\;dx \\ & = \dfrac{1}{2} \dfrac{f(x)^4}{4} \\ & = \dfrac{(x^2 + 1)^4}{8} + C. \end{align*}

Observe that once we identify the function $f$ and the power ${\displaystyle r\text{,}}$ we take the derivative ${\displaystyle f^{\prime }\text{,}}$ and modify the integrand function with a constant to fit the form ${\displaystyle f{\left(x\right)}^r{f}^{\prime }\left(x\right)\text{.}}$ If we cannot modify it with a constant, then we must use another method of integration.

Example 6.12. In the integral ${\displaystyle \int \sin x\cos xdx\text{,}}$ we identify ${\displaystyle f\left(x\right)=\sin x}$ and ${\displaystyle r=1}$; thus, $f^{\prime }\left(x\right)=\cos x$ and

${\displaystyle \int \sin x\cos xdx}={\displaystyle \int f(x)f^{\prime }(x)dx}={\displaystyle \frac{f{(x)}^2}{2}}={\displaystyle \frac{{\sin }^{2}x}{2}}+C$.

We can also identify ${\displaystyle f\left(x\right)=\cos x}$ and ${\displaystyle r=1}$; thus, $f^{\prime }\left(x\right)=-\sin x$ and

${\displaystyle \int \sin x\cos xdx}=-{\displaystyle \int f(x)f^{\prime }(x)dx}=-{\displaystyle \frac{f{(x)}^2}{2}}=-{\displaystyle \frac{{\cos }^{2}x}{2}}+C$.

So, we have found two antiderivatives of the function $\sin x\cos x$. However, if we use a trigonometric identity we see that

${\displaystyle \frac{{\sin }^{2}x}{2}}={\displaystyle \frac{1-{\cos }^{2}x}{2}}=-{\displaystyle \frac{{\cos }^{2}x}{2}}+{\displaystyle \frac{1}{2}}$.

Therefore, the antiderivatives are equal except for a constant, as we said earlier. The promised proof is given next.

Theorem 6.2. If two functions ${\displaystyle f_1}$ and ${\displaystyle f_2}$ are the antiderivative of a function ${\displaystyle f\text{,}}$ then ${\displaystyle f_1=f_2+K}$ for some constant ${\displaystyle K\text{.}}$

Proof. Since

${\displaystyle {\displaystyle \frac{d}{dx}}{f}_1\left(x\right)-f_2\left(x\right)=f_1^{\prime }\left(x\right)-f_2^{\prime }\left(x\right)=f\left(x\right)-f\left(x\right)=0,}$

by Example 6.1(a), we know that ${\displaystyle f_1\left(x\right)-f_2\left(x\right)=K\text{,}}$ for some constant $K$; and therefore,

${\displaystyle f_1\left(x\right)=f_2\left(x\right)+K\text{.}}$ Q.E.D

Example 6.13. In the integral ${\displaystyle \int \tan x{\sec }^{2}xdx\text{,}}$ we identify ${\displaystyle f\left(x\right)=\tan x}$ and ${\displaystyle r=1\text{,}}$ because ${\displaystyle f^{\prime }\left(x\right)={\sec }^{2}x\text{.}}$ Thus,

${\displaystyle \int \tan x{\sec }^{2}xdx={\displaystyle \frac{{\tan }^{2}x}{2}}+C}$.

Example 6.14. To identify the function $f$ and the power $r$ in the integral

${\displaystyle \int {\displaystyle \frac{x}{\sqrt{3x^2-5}}}dx,}$

we observe that

${\displaystyle {\displaystyle \frac{x}{\sqrt{3x^2-5}}}=x{\left(3x^2-5\right)}^{-1∕2}\text{.}}$

Hence, ${\displaystyle f\left(x\right)=3x^2-5}$ and ${\displaystyle r=-{\displaystyle \frac{1}{2}}\text{.}}$

Since ${\displaystyle f^{\prime }\left(x\right)=6x\text{,}}$ we proceed as follows:

\begin{align*} \int {\dfrac{x}{{\sqrt {3{x^2} - 5} }}\,} dx & = {\int {x(3{x^2} - 5)} ^{ - 1/2}}\,dx \\ & = \dfrac{1}{6}{\int {(3{x^2} - 5)} ^{ - 1/2}}\,(6x)\,dx \\ & = \dfrac{1}{6}\dfrac{{{{(3{x^2} - 5)}^{1/2}}}}{{1/2}} \\ & = \dfrac{{\sqrt {3{x^2} - 5} }}{3} + C{\text{.}} \end{align*}

Example 6.15. The function raised to a power in the integral

${\displaystyle \int {\cos }^3\left(4x\right)\sin \left(4x\right)dx}$

is

${\displaystyle f\left(x\right)=\cos \left(4x\right)\text{.}}$

Thus ${\displaystyle r=3\text{,}}$ and ${\displaystyle f^{\prime }\left(x\right)=-4\sin \left(4x\right)\text{.}}$

Then,

\begin{align*} \int \cos^3(4x)\sin (4x) \, dx & = -\dfrac{1}{4} \int \cos^3(4x)(-4\sin(4x))\,dx \\ & = -\dfrac{1}{4}\dfrac{\cos^4(4x)}{4} + C \\ & = -\dfrac{\cos^4(4x)}{16} + C {\text{.}} \end{align*}

Example 6.16. In this example, we ask you to identify the function $f$ and the power ${\displaystyle r\text{.}}$

\begin{align*} \int \dfrac{2x^2}{\sqrt{4x^3 + 4}}\,dx & = \int (4x^3 + 4)^{-1/2}\,(2x^2)\,dx \\ & = \dfrac{1}{6} \int (4x^3 + 4)^{-1/2}\,(12x^2)\,dx \\ & = \dfrac{(4x^3 + 4)^{1/2}}{3} + C {\text{.}} \end{align*}