Question

Help please!

Solve the definite integral.

integral of xsinx^2dx from 0 to sqrt 4x

Answer 1

u = x^2

Answer 2

I'm assuming that the integral is:
\[\int\limits x Sin(x^2)dx\]

Answer 3

so it would be integral of sin(u) du

Answer 4

There should be an additional constant in there, but close :)

Answer 5

ok so it would be the integral of sin which is cos and that becomes cos(u) +c and that then becomes.... cos(x^2)+c

Answer 6

\[\int\limits x Sin(x^2)dx = \frac{1}{2}\int\limits \sin(u)du = \frac{1}{2}[-\cos(u)]+C\]

Since you have bounds, we don't really care about the C, though.

Answer 7

ok so sqrt 4pi would be first :)

Answer 8

Well, your bounds were for x. You either need to find what u is when x = those things, and use those as your new bounds, or substitute back in for u.

Answer 9

ok so it would be 1/2 -cos (x^2) since u is x^2

Answer 10

Just to be clear, the 1/2 is multiplying the (-cos(u)). Rewritten for clarity:

\[\frac{-1}{2}Cos(u)\]

Answer 11

yeah sorry that is what i mean :)

Answer 12

Right. and u = x^2, so you plug that in, then evaluate at the given bounds.

Answer 13

so then it is sqrt of 4pi first which when squared becomes 4pi and -cos(4pi) = -1 and divided by 2 = -0.5

Answer 14

subtract -cos(0)/2 from that, which equals -0.5 and the answer is 0 :)

Answer 15

Looks good to me :)

Answer 16

yay! :D thank you!

Answer 17

My pleasure :)