Question

.... do not enter!!

Answer 1

just doin homework :)

Answer 2

AMISTRE I NEED HELP!!

Answer 3

cant help right at the moment ...

Answer 4

:/ Kayy!

Answer 5

is sulfur allowed?

Answer 6

Okay . thanks .

Answer 7

no sulfur :) unless its catching my typos ...

Answer 8

but thats not fair
whenever i smell you i smell sulfur

Answer 9

Sorry. :(

Answer 10

\[\Large
\begin{align}
4)\ \int {xe^{-x}}\ dx&=-xe^{-x}-\int e^{-x}dx\\
&=-xe^{-x}+-e^{-x}dx\\
&=-e^{-x}(x+1)+C\\
\end{align}
\]
\[\Large
\begin{align}
5)\ \int {re^{r/2}}\ dr&=2re^{r/2}-\int e^{r/2}dr\\
&=2re^{r/2}-\int 2e^{r/2}dr\\
&=2re^{r/2}-4e^{r/2}+C\\
\end{align}
\]

Answer 11

mistake on number 4 i believe

Answer 12

i see the dx drs that i can erase ... but other than that

Answer 13

right

Answer 14

but i accounted for it in the end

Answer 15

oops i made a type-o too

Answer 16

\[-xe^{-x}-\int\limits\limits_{}^{}-e^{-x} dx\]

Answer 17

all that results is the signs swap out and we get back to it in the end right?

Answer 18

ok i just would have taken off a point for not shown your work correctly lol

Answer 19

\[\Large
\begin{align}
6)&\ \int {t\ sin(2t)}\ dt\\\\
&=\frac{-t\ cos(2t)}{2}-\int \frac{-cos(2t)}{2}dt\\\\
&=\frac{-t\ cos(2t)}{2}-\frac{-sin(2t)}{4}\\\\
&=\frac{-t\ cos(2t)}{2}+\frac{sin(2t)}{4}+C\\
\end{align}\]

Answer 20

\[\Large
\begin{align}
10)&\ \int {sin^{-1}(x)}\ dx\\\\
&=x\ sin^{-1}(x)-\int\frac{x}{\sqrt{1-x^2}}\ dx\\\\
&=x\ sin^{-1}(x)+\int\frac{2x}{2\sqrt{1-x^2}}\ dx\\\\
&=x\ sin^{-1}(x)+\sqrt{1-x^2}+C\\\\
\end{align}\]

Answer 21

\[\Large
\begin{align}
19)&\ \int_{0}^{\pi} {t\ sin(3t)}\ dt\\\\
&=\frac{-t\ cos(3t)}{3}-\int \frac{-cos(3t)}{3}dt\\\\
&=\left.\frac{-t\ cos(3t)}{3}+ \frac{sin(3t)}{9}\ \right|_{0}^{\pi}\\\\
&=\frac{-\pi\ cos(3\pi)}{3}+ \frac{sin(3\pi)}{9}\\\\
&+\frac{0\ cos(3(0))}{3}- \frac{sin(3(0))}{9}\\\\
&=\frac{\pi}{3}\\\\

\end{align}\]

Answer 22

\[\Large
\begin{align}
23)&\ \int_{1}^{2} {\frac{ln(x)}{x^2}}\ dx\\\\
&=\frac{-ln(x)}{x}+\int \frac{1}{x^2}dx\\\\
&=\frac{-ln(x)}{x}-\left.\frac{1}{x}\ \right|_{1}^{2}\\\\
&=\frac{-ln(2)}{2}-\frac{1}{2}\\\\
&+\frac{ln(1)}{1}+\frac{1}{1}\\\\
&=\frac{-ln(2)}{2}+\frac{1}{2}\\\\

\end{align}\]

Answer 23

\[\Large
\begin{align}
34)&\ \int {t^3\ e^{-t^2}}\ dt\\\\
&=-\frac12\int {t^2(-2t\ e^{-t^2})}\ dt\\\\
&=-\frac12\left({t^2e^{-t^2}+\int -2te^{-t^2}}\ dt\right)\\\\
&=-\frac12\left({t^2e^{-t^2}+e^{-t^2}}\ \right)\\\\
&=-\frac{e^{-t^2}}{2}({t^2+1})+C\\\\
\end{align}\]

Answer 24

\[\Large
\begin{align}
43)a.&\ \int sin^2(x)\ dx\\\\
&=\frac{x}{2}-\frac{sin(2x)}{4}+C\\\\
&=-\frac{1}{n}sin^{n-1}(x)cos(x)+\frac{n-1}{n}\\\\
&\int sin^{n-2}(x)\ dx\\\\
&=-\frac{1}{2}sin^{2-1}(x)cos(x)+\frac{2-1}{2}\\\\
&\int sin^{2-2}(x)\ dx\\\\
&=-\frac{1}{2}sin(x)cos(x)+\frac{1}{2}\int \ dx\\\\
&=-\frac{1}{2}sin(x)cos(x)+\frac{1}{2}x\\\\
&=\frac{x}{2}-\frac{1(2)}{2(2)}sin(x)cos(x)\\\\
&=\frac{x}{2}-\frac{sin(2t)}{4}+C\\\\

\end{align}\]
\[\Large
\begin{align}
43)b.&\ \int sin^4(x)\ dx\\\\
&=-\frac{1}{4}sin^{3}(x)cos(x)+\frac{3}{4}\\\\
&\int sin^{2}(x)\ dx\\\\

&=-\frac{1}{4}sin^{3}(x)cos(x)+\frac{3}{4}\\\\
&\left(\frac{x}{2}-\frac{sin(2t)}{4}\right)+C\\\\
\end{align}\]

Answer 25

\[\Large
\begin{align}
30)&\ \int_{0}^{1} {\frac{r^3}{\sqrt{4+r^2}}}\ dr\\\\
&=\int_{0}^{1} {\frac{2r}{2\sqrt{4+r^2}}}\ r^2\ dr\\\\
&=r^2\sqrt{4+r^2}-\int 2r \sqrt{4+r^2}\ dr\\\\
&=r^2\sqrt{4+r^2}-\left.\frac {2}{3}(4+r^2)^{3/2}\ \right|_{0}^{1}\\\\
&=\sqrt{5}-\frac {2}{3}(5)^{3/2}+\frac {2}{3}(4)^{3/2}\\\\
&=\sqrt{5}-\frac {2(5)}{3}\sqrt{5}+\frac {2(8)}{3}\\\\
&=\frac{3}{3}\sqrt{5}-\frac {10}{3}\sqrt{5}+\frac {16}{3}\\\\
&=\frac{16-7\sqrt{5}}{3}
\end{align}\]

Answer 26

how are you planning on turning it in?

Answer 27

yep