Question

\[\int^{2}_{-1} xe^{6x}dx\]

Answer 1

This time again?\[ u = x\\dv = e^{6x}dx\\du = dx \\ v = {e^{6x} \over 6} + c\]

Answer 2

That + c must not be there... sorry

Answer 3

\[\int_{-1}^{2} xe^{6x}dx = {xe^{6x} \over 6} - \int_{-1}^{2} {e^{6x} \over 6}dx \]

Answer 4

What should I do now?

Answer 5

Take the constant out:
= (1/6 ) | e^(6x) x - e^(6x) |

Answer 6

Didn't think of that... lol

Answer 7

And that step was wrong.... :|

Answer 8

\[\int_{-1}^{2} xe^{6x}dx = {xe^{6x} \over 6} - \int_{-1}^{2} {e^{6x} \over 6}dx \]
\[\int_{-1}^{2} xe^{6x}dx = {xe^{6x} \over 6} - {1 \over 6}\int_{-1}^{2} {e^{6x}}dx \]

Answer 9

Wait... \[ \implies \left.\left(e^6x^3 \over 3 \right)\right|^{2}_{-1} =\left.{xe^{6x} \over 6}\right|^{2}_{-1} - {1 \over 6}\left.\left({e^{6x} \over 6} \right)\right|^{2}_{-1}\]

Answer 10

That's really long, but am I right till this step? @Chlorophyll

Answer 11

x shouldn't be exist any more!

Answer 12

Can you tell me what'd be correct?

Answer 13

= (1/6) [ 2e^12 + e^-6 ] - (1/36) [ e^12 + e^-6 ]

Answer 14

Hmm, I really really give up... this is all too hard and tedious! The good thing is that I got the basic idea :)

Answer 15

Watch the link I just sent to you. It's much more interesting stuff!
Not to be hurry work on the definite integral when you're still not strong with indefinite one!

Answer 16

OK :)