Question

clubhouse, no chemists allowed :)

Answer 1

doing me homewirk ...

Answer 2

\[\left. \frac{x^2}{2}\right| _a^b\]

Answer 3

You guys are the best!

Answer 4

\[\Large

\begin{array}l
20)\ \int_{-2}^{5}6\ dx&\left.=6x\ \right|_{-2}^{5}\\\\
&=6(5-(-2)) = 42\\\\
\end{array}
\]

\[\Large

\begin{array}l

22)\ &\int_{0}^{1}(1+\frac{1}{2}u^4-\frac{2}{5}u^9)\ du\\\\
&=u+\frac{1}{10}u^5-\left.\frac{2}{50}u^{10}\ \right|_{0}^{1}\\\\
&=1+\frac{1}{10}-\frac{2}{50}\\\\
&=\frac{50}{50}+\frac{5}{50}-\frac{2}{50}\\\\
&=\frac{53}{50}\\\\
\end{array}
\]

\[\Large

\begin{array}l

24)\ \int_{1}^{8}\sqrt[3]{x}\ dx&=\left.\frac{3}{4}x^{4/3}\ \right|_{1}^{8}\\\\
&=\frac{3}{4}(8^{4/3}-1^{4/3})\\\\
&=\frac{3}{4}(8(2)-1)\\\\
&=\frac{3}{4}(15)\\\\
&=\frac{45}{4}\\\\
\end{array}
\]

\[\Large

\begin{array}l
26)\ \int_{\pi}^{2\pi}\ cos(t)\ dt&=\left.sin(t)\ \right|_{\pi}^{2\pi}\\\\
&=sin(2\pi)-sin(\pi)\\\\
&=0
\end{array}\]

\[\Large
\begin{array}l
28)\ \int_{0}^{1} (3+x\sqrt{x})\ dx&=\left.{3x+\frac{2}{5}x^{5/2}}\ \right|_{0}^{1}\\\\
&=3+\frac{2}{5}\\\\
&=\frac{17}{5}
\end{array}\]


\[\Large
\begin{array}l
30)\ \int_{0}^{2}(y-1)(2y+1)\ dy&=\left.{\frac{2}{3}y^3-\frac{1}{2}y^2-y}\ \right|_{0}^{2}\\\\
&=\frac{2}{3}2^3-\frac{1}{2}2^2-2\\\\
&=\frac{16}{3}-\frac{4}{2}-2\\\\
&=\frac{16}{3}-\frac{12}{3}\\\\
&=\frac{4}{3}\\\\
\end{array}\]


\[\Large
\begin{array}l
32)\ \int_{0}^{\pi/4}sec(t)\ tan(t)\ dt&=\left.{sec(t)}\ \right|_{0}^{\pi/4}\\\\
&=sec(\pi/4)-sec(0)\\\\
&=\sqrt{2}-1\\\\
\end{array}\]
check for continuity ...

\[\Large
\begin{array}l
38)\ \int_{0}^{1}\frac{4}{t^2+1}\ dt&=\left.{4\ tan^{-1}(t^2+1)}\ \right|_{0}^{1}\\\\
&=4(tan^{-1}(1^2+1)-tan^{-1}(1))\\\\
&=4\ tan^{-1}(2)-\pi\\\\
\end{array}\]

\[\Large
\begin{array}l
39)\ \int_{-1}^{1}\ e^{u+1}\ du&=\left.{e^{u+1}}\ \right|_{-1}^{1}\\\\
&=e^{1+1}-e^{-1+1}\\\\
&=e^{2}-e^{0}\\\\\
&=e^{2}-1\\\\\
\end{array}\]


\[\Large
\begin{array}l
40)\ \int_{1}^{2}\frac{4+u^2}{u^3}\ &=\left.{-\frac{2}{u^2}+ln|u|}\ \right|_{1}^{2}\\\\
&=-\frac{2}{2^2}+ln(2)+\frac{2}{1^2}-ln(1)\\\\
&=-\frac{1}{2}+ln(2)+2\\\\
&=\frac{3}{2}+ln(2)\\\\
\end{array}\]

gonna need to know how to do a piecewise latex for the last one
\[\Large
\begin{array}l
)\ \int_{}^{}\ &=\left.{}\ \right|_{}^{}\\\\
\end{array}\]

Answer 5

and type in a du that i missed

Answer 6

\[f(x) = |x+4| = \left\{\begin{array}{rcc}
x + 4 & \text{if} & x \geq -4 \\
- x - 4& \text{if} & x < -4
\end{array}
\right.
\]

Answer 7

err amistre can you plz argue that the integral of cos(x) = sin(x) oO

Answer 8

the arguement is simply the reversal of the derivative
\[\frac{d}{dx}sin(x)=cos(x)\]

Answer 9

\[\Large

42)\ \int_{-2}^{2}f(x)\ dx = \left\{\begin{array}{lcc}
\ 2 &; \text{if} &-2\le x\le 0\\\\
\ 4-x^2&; \text{if} & 0\lt x\le 2
\end{array}
\right.

\]

Answer 10

hmmm

Answer 11

\[\Large
\begin{array}l
A=\int_{-2}^{0} 2\ dx&=\left.2x\ \right|^{0}_{-2}\\\\
&=2(0+2)\\\\
&=4
\end{array}
\]

\[\Large
\begin{array}l
B=\int_{0}^{2} 4-x^2\ dx&=\left.{4x-\frac{1}{3}x^3}\ \right|^{2}_{0}\\\\
&=4(2)-\frac{1}{3}(2)^3\\\\
&=8 - \frac{8}{3}\\\\
&=\frac{16}{3}\\\\
&\\\\
&\\\\
&\\\\
A+B&=4 + \frac{16}{3}\\\\
&=\frac{28}{3}
\end{array}
\]

Answer 12

how many questions you have at total? :D

Answer 13

thats it for the first page :)

Answer 14

I found that I can copy from the paint and paste directly into word ...

Answer 15

14 more on the other page :)

The last times i did this i used the normal size latex stuff and everything tends to look the same. With this size you can see more detail

Answer 16

yea you right :D

Answer 17

nice questions there :D

Answer 18

\[\Large
\begin{array}l
6)\ \int\ ({x^{3/2} +x^{2/3}})\ {dx}={\frac{2}{5}x^{5/2}+\frac{3}{5}x^{5/3} }+C\\\\

\end{array}\]
\[\Large
\begin{array}l
8)\ \int\ {(y^3+1.8y^2}&{-2.4y) }\ {dy}\\\\
&={\frac{1}{4}y^4+.6y^3-1.2y^2}+C\\\\

\end{array}\]
\[\Large
\begin{array}l
10)\ \int\ {v(v^2+2)^2 }\ {dv} &=\int\ {(v^5+4v^3+4v)} \ dv\\\\
&=\frac{1}{6}v^6+v^4+2v^2+C

\end{array}\]
\[\Large
\begin{array}l
12)\ \int\ {\left({x^2+1+\frac{1}{x^2+1}}\right) }\ {dx}\\\\
={\frac{1}{3}x^3+x+tan^{-1}(x^2+1)}+C\\\\

\end{array}\]
\[\Large
\begin{array}l
14)\ \int\ {(csc^2(t)-2e^t) }\ {dt}={-cot(t)-2e^t }+C\\\\

\end{array}\]
\[\Large
\begin{array}l
16)\ \int\ {sec(t)\ [sec(t)+tan(t)] }\ {dt}\\\\
={\int\ {sec^2(t)+sec(t)\ tan(t) }\ {dt} }\\\\
={tan(t)+sec(t)}+C\\\\

\end{array}\]
\[\Large
\begin{array}l
18)\ \int\ {\frac{sin(2x)}{sin(x)} }\ {dt}\\\\
={\int\ {\frac{2\ sin(x)\ cos(x)}{sin(x)} }\ {dt} }\\\\
={\int\ {2\ sin(x)}\ {+C} }\\\\

\end{array}\]

Answer 19

\[\Large
\begin{array}l
18)\ \int\ {\frac{sin(2x)}{sin(x)} }\ {dx}\\\\
={\int\ {\frac{2\ sin(x)\ cos(x)}{sin(x)} }\ {dx} }\\\\
={\int\ {2\ sin(x)}\ {+C} }\\\\

\end{array}\]

Answer 20

feel free to check the work as well;)

Answer 21

i se a few typos that i took care of in the print tho

Answer 22

kk :D but i am not good at doing calculus with trigonometry, have to work on it before it beings a problem for me ...