Question

I'm beginning to understand integration... anyone mind giving me something to try out...you guys got questions of which u know the answers... i wanna try them too so let me have them and let me know if i make a mistake... :/

Answer 1

integrate cos 3x with respect to x

Answer 2

\[\int_0^{\infty} 2^t e^t dt\]

Answer 3

jim: 1/3 sin3x + c

Answer 4

yup

Answer 5

\[\int\limits{24x \ln( 6x^2) dx}\]

Answer 6

want an improper??

Answer 7

\[\huge \int_{1}^{2}\frac{cosh(lnt)}{t}dt\]

Answer 8

try this
\[\int\limits_{?}^{?} 1/x\]

Answer 9

dx

Answer 10

i was trying to solve lana's question but i'm stuck at solving the natural log... need help

Answer 11

get mine?

Answer 12

by parts saso

Answer 13

imran i can't do ur question... the very first one... and lana: int dx = x

Answer 14

anyone?

Answer 15

no lol there was a dx missing in that persons integral

Answer 16

\[\int\limits_{0}^{1}2x \div(x ^{2}+3) \]

Answer 17

but is the integration of dx = x? was i right?

Answer 18

yup:D

Answer 19

what about the intergration of 1/x?

Answer 20

ln x

Answer 21

btw though i don't know how to solve integrals with a linear function involving fractions where the variable is a part of the denominator... like jimmy's last question... i'm going to learn that later today or better still someone could explain it to me now...

Answer 22

try mine. It's just algebra

Answer 23

imran i'm going to try and solve urs on a sheet of paper first before trying to type it out so you'd have to wait for a bit :)

Answer 24

for jimmys you can is U sub

Answer 25

so u = \[u = x^{2}+3\]
Du = 2x

Answer 26

i think thats it im not sure i jus got out of a 3 hour exam and am tired lol

Answer 27

imran the answer is 2

Answer 28

right?

Answer 29

\[\int_0^{\infty} 2^t e^t dt=\infty\]

Answer 30

\[\[2^t= e^{t *log(2)}\]

\[\int_0^{\infty} e^{t *log(2)} * e^t dt\]

\[\int_0^{\infty} e^{t *log(2)+t}dt \]

\[\int_0^{\infty} e^{t (*log(2)+1)} \]

Answer 31

I know we could have used comparison test, but this is for any limits

Answer 32

\[2^t e^t \ge1 \text{ for }t\ge 0\] thus

\[\int_0^{\infty} 2^t e^t dt\ge \int_0^{\infty} 1 dt=\infty\]

Answer 33

\[\int\limits_{0}^{\infty}2^{t}e ^{t}dt=\left[ ((2^{t+1})/t+1)e ^{t} \right]_{0}^{\infty}\]

Answer 34

I would just write \[2^te^t\]as\[(2e)^t\]

Answer 35

use \[\int a^tdt=\frac{a^t}{\ln(a)}+c\]

Answer 36

I got this out of finding laplace transform of 2^t

Answer 37

omg i made a huge mistake lol

Answer 38

the 2 was a base to t, not a coefficient so i integrated so wrongly

Answer 39

facepalm

Answer 40

zarkon i haven't seen ur formula before though... there's a lot i'm yet to learn :P i didn't expect this level of questions lol but it was worth seeing what the future beholds for me