# C.3 Integrals

C.3 Integrals

As with the review of Derivatives, it would be challenging to include a full review of Integrals. In this review, we try to include the most common integrals and rules used in STAT 414. There are many helpful websites as texts out there to help you review. We have provided links to Khan Academy for you to take a look at if you have difficulty recalling these methods.

For a function, $$f(x)$$, its indefinite integral is:

$\int f(x)\; dx=F(x)+C, \qquad \text{where } F^\prime(x)=f(x)$

We provide a short list of common integrals and rules that are used in STAT 414. It is important to have a lot of practice and keep these skills fresh.

1. Common Integrals and Rules

1. $$\int_a^a f(x)dx=0$$
2. $$\int_a^b f(x)d(x)=-\int_b^a f(x)d(x)$$
3. $$\int x^rdx=\frac{x^{r+1}}{r+1}+C$$
4. The Fundamental Theorem of Calculus: Let $$f$$ be integrable on $$[a,b]$$ and let $$F$$ be any antiderivative of $$f$$ there. Then, $$\int_a^b f(x)d(x)=F(b)-F(a)$$. (FTC, Khan Academy)
5. $$\int x^n dx=\dfrac{1}{n+1}x^{n+1}+C, \;\;n\neq(-1)$$
6. $$\int \dfrac{1}{x}dx=\ln |x| +C$$
7. $$\int e^x dx=e^x +C$$
2. Integration Using Substitution: Let $$g$$ have a continuous derivative on $$[a,b]$$ and let $$f$$ be continuous on the range of $$g$$. Then

$\int_a^b f\left(g(x)\right)g^\prime(x)dx=\int_{g(a)}^{g(b)}f(u)du$

where $$u=g(x)$$. (u-Substitution, Khan Academy)

3. Integration by Parts

$\int_a^b udv=\left[uv\right]_a^b-\int_a^b vdu$.

## Example C.3.1

Integrate the following function from 0 to $t$

$f(x)=\dfrac{2}{1000^2}xe^{-(x/1000)^2}$

$\int_0^t \frac{2}{1000^2}xe^{-(x/1000)^2} dx\label{eqn1}$

Let $$u=\left(\frac{x}{1000}\right)^2$$. Then $$du=\frac{2}{1000^2}xdx$$. The equation becomes...

\begin{align*} &= \int_0^{\left(\frac{t}{1000}\right)^2} e^{-u}du =-e^{-u}|_{0}^{\left(\frac{t}{1000}\right)^2}\\ &= -e^{-\left(\frac{t}{1000}\right)^2}-(-1)=1-e^{-\left(\frac{t}{1000}\right)^2}. \end{align*}

## Example C.3.2

Integrate the following:

$\int_0^5 x^2e^{-x}dx$

Let us begin by setting up integration by parts. Let

\begin{align*} & u=x^2 \qquad dv=e^{-x}dx\\ & du=2xdx \qquad v=-e^{-x} \end{align*}

Then

\begin{align*} uv|_0^5-\int_0^5 vdu &=-x^2e^{-x}|_0^5+2\int_0^5xe^{-x}dx\\ &= -x^2e^{-x}|_0^5+2\left[-xe^{-x}|_0^5+\int_0^5 e^{-x}dx\right]\\ &= -x^2e^{-x}|_0^5+2\left[-xe^{-x}|_0^5-e^{-x}|_0^5\right]\approx 1.75 \end{align*}

## Example C.3.3

Integrate the following from $$-\infty$$ to $$\infty$$.

$f(y)=\frac{1}{2}e^{-|y|+ty}, \;\; \text{ for } -\infty<y<\infty$

\begin{align*}
\int_{-\infty}^{\infty} \frac{1}{2} e^{ty-|y|}dy &= \int_{-\infty}^0 \frac{1}{2}e^{y+ty}dy+\int_0^{\infty} \frac{1}{2}e^{-y+ty}dy\\
& = \int_{-\infty}^0 \frac{1}{2}e^{y(1+t)}dy+\int_0^{\infty} \frac{1}{2}e^{-y(1-t)}dy\\
& = \frac{1}{2(1+t)}+ \frac{1}{2(1-t)}=\frac{1}{2}\left(\frac{1-t+t+1}{(1+t)(1-t)}\right)\\
& =\frac{1}{(1-t)(1+t)}
\end{align*}

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