Question

help with integrals...

Answer 1

http://i50.tinypic.com/bgtk7n.jpg

Answer 2

what is it you are having trouble with?

Answer 3

on number 28 would you just plug -1 and 1 in?

Answer 4

for what is that for?

Answer 5

what do you mean

Answer 6

i just want to clarify things man

Answer 7

on number 28 it says that the derivative of h(x) is k(x)

Answer 8

the integral is the opposite of the derivative, so the integral of k(5x) would be h(5x)

Answer 9

then substitute -1 and 1 for x?

Answer 10

yes

Answer 11

that is right @Peter14 . i know man. and the answer is 0, right?!

Answer 12

no, it's answer a.

Answer 13

is that so? then, i should be careful, i guess :)

Answer 14

for 26. would i need to do u substitution?

Answer 15

i think so

Answer 16

@Peter14 shouldn't 28 be E, \[\frac{ 1 }{ 5 } h(5) - \frac{ 1 }{ 5 } h(-5)\]

Answer 17

how?

Answer 18

you have a point @agent0smith . i still get a little confuse with the given

Answer 19

maybe, I haven't done calculus since last year.

Answer 20

Let's imagine k(x) = cosx.

Then k(5x) = cos(5x).

\[\large \int\limits_{-1}^{1} k(5x) dx = \int\limits_{-1}^{1} \cos(5x) dx = \left[ \frac{ \sin(5x) }{ 5} \right]_{-1}^{1}\]

Answer 21

oh, yes because you do u-substitution to do the integral. I see.

Answer 22

aha! that's it. you guys are great

Answer 23

"for 26. would i need to do u substitution?" No, you won't actually need to integrate. The derivative kinda cancels out the integral (I'm a little rusty on explaining this part, someone else may do a better job)

it's essentially saying: \[\large f'(x) = \frac{ d }{ dx } \int\limits_{1}^{x} t(t^3+1) dt\]

Answer 24

\[\large f'(x) = \frac{ d }{ dx } \int\limits\limits_{1}^{x} t(t^3+1)^{\frac{ 3 }{x }} dt = x(x^3+1)^{\frac{ 3 }{ 2 }}\]
I think that's right, since the derivative of the x is just 1 in this case. Hopefully someone will correct me if I'm wrong. Now you just put in x=2.

Answer 25

thank you... so since its asking for the derivative of an integral it cancels out and you just plug in 2...

Answer 26

In this case, yes because the upper limit is just x. If you had 5x instead, then you have to differentiate that as well because you'd be plugging that in for t after integrating, and then differentiating w.r.t. x, but... I just can't really find an easy way to explain that.