Question

...... more homework :)

Answer 1

the other place got bogged down

Answer 2

i miss homework :)

Answer 3

homework!!!!

Answer 4

\[\Large
\begin{array}l
26)\ \int_{0 }^{4 }{(2v+5)(3v-1) }\ {dv}\\\\=\int_{0 }^{4 } {(6v^2+13v -5) }\ {dv}\\\\
=\left.{2v^3+\frac{13}{2}v^2-5v }\ \right|_{0 }^{4 }\\\\
={2(4)^3+\frac{13}{2}(4)^2-5(4) }\\\\
={2(64)+\frac{13(2)}{1}-20 }\\\\
={128+26-20 }\\\\
={134}\\\\
\end{array}\]

\[\Large
\begin{array}l
28)\ \int_{0 }^{9 }\ {\sqrt{2t} }\ {dt }&=\left.{\frac{2\sqrt{2}}{3}t^{3/2} }\ \right|_{0 }^{9 }\\\\
&={\frac{2\sqrt{2}}{3}(9)^{3/2} }\\\\
&={18\sqrt{2} }\\\\
\end{array}\]

\[\Large
\begin{array}l
30)\ \int_{1 }^{2 }\ {\frac{y+5y^7}{y^3} }\ {dy }&=\left.{-\frac{1}{y}+y^5 }\ \right|_{ 1}^{2 }\\\\
&=-\frac{1}{2}+(2)^5 -(-1+1)\\\\
&=-\frac{1}{2}+32\\\\
&=\frac{63}{2}\\\\
\end{array}\]

\[\Large
\begin{array}l
32)\ \int_{0 }^{5 }\ {(2e^x+4\ cos(x)) }\ {dx }=\left.{2e^x+4\ sin(x) }\ \right|_{0 }^{5 }\\\\
=2e^5+4\ sin(5)-(2e^0+4\ sin(0))\\\\
=2(e^5+2\ sin(5)-1)
\end{array}\]

\[\Large
\begin{array}l
)\ \int_{ }^{ }\ { }\ {d }&=\left.{ }\ \right|_{ }^{ }\\\\
\end{array}\]

Answer 5

but the funniest is not to solve your homework then go to your teacher and say "my cat ate my homework" :D

Answer 6

perhaps :)

Answer 7

i still can't understand integrals with trigonometry or derivatives with trigonometry, it makes me confused ...

Answer 8

amistre you are saving all this ....right? making a cloud

Answer 9

\[ \Large
\begin{array}l
34)\ \int_{1 }^{ 9}\ {\frac{3x-2}{x^{1/2}} }\ {dx}=\ \int_{1 }^{ 9}\ {\frac{3x}{x^{1/2}}-\frac{2}{x^{1/2}} }\ {dx}\\\\
=\left.{x^{3/2}-4x^{1/2} }\ \right|_{1 }^{9 }\\\\
=(9)^{3/2}-4(9)^{1/2}-((1)^{3/2}-4(1)^{1/2})\\\\
=27-12-(1-4)\\\\
=18\\\\
\end{array}\]

\[ \Large
\begin{array}l
36)\ \int_{\pi/4 }^{ \pi/3}\ {sec(t)tan(t) }\ {dt}&=\left.{sec(t) }\ \right|_{\pi/4 }^{\pi/3 }\\\\
&=sec(\pi/4)-sec(\pi/3)\\\\
&=\sqrt{2}- 2\\\\
\end{array}\]


\[ \Large
\begin{array}l
38)\ \int_{0 }^{\pi/3 }\ {\frac{sin(t)+sin(t)tan^2(t)}{sec^2(t)} }\ {dt }\\\\
=\ \int_{0 }^{\pi/3 }\ {\frac{sin(t)(1+tan^2(t))}{sec^2(t)} }\ {dt }\\\\
=\ \int_{0 }^{\pi/3 }\ {\frac{sin(t)sec^2(t)}{sec^2(t)} }\ {dt }\\\\
=\left.{-cos(t) }\ \right|_{0 }^{\pi/3 }\\\\
=-cos(\pi/3)+cos(0)\\\\
=-\frac{1}{2}+1\\\\
=\frac{1}{2}
\end{array}\]

Answer 10

yeah, im clouding lol

Answer 11

nah, its just easier for me to type it up and have it all purty this way

Answer 12

:D

Answer 13

and were done .. yay!!

Answer 14

finallY!!! after hours of work :D

Answer 15

now lets hope that its all correct ...

Answer 16

ahaha :D well i can say that the integrals u solved which not includes trigonometry are correct, but for somes includes trigonometry i can2t say anything :(

Answer 17

when integrals or derivatives calculations includes trigonometry, it causes me to get confused -.-'

Answer 18

Thats some hard core math.

Answer 19

well, i found some mathical errors .... in 2 of them

\[\Large
\begin{array}l
26)\ \int_{0 }^{4 }{(2v+5)(3v-1) }\ {dv}\\\\=\int_{0 }^{4 } {(6v^2+13v -5) }\ {dv}\\\\
=\left.{2v^3+\frac{13}{2}v^2-5v }\ \right|_{0 }^{4 }\\\\
={2(4)^3+\frac{13}{2}(4)^2-5(4) }\\\\
={2(4)^3+\frac{13}{2}(16)-5(4) }\\\\
={2(64)+13(8)-20 }\\\\
={128+104-20 }\\\\
={212}\\\\
\end{array}\]
\[
\Large
\begin{array}l
34)\ \int_{1 }^{ 9}\ {\frac{3x-2}{x^{1/2}} }\ {dx}=\ \int_{1 }^{ 9}\ {3x^{1/2}-2x^{-1/2}}\ {dx}\\\\
=\left.{2x^{3/2}-4x^{1/2} }\ \right|_{1 }^{9 }\\\\
=2(9)^{3/2}-4(9)^{1/2}-(2(1)^{3/2}-4(1)^{1/2})\\\\
=2(27)-12-(2-4)\\\\
=54-12-(-2)\\\\
=44\\\\
\end{array}\]

Answer 20

are you turning in like this?