Question

I can't get my answers right, can someone please check my work?

Answer 1

What's the work?

Answer 2

Here's the first question:

Answer 3

Here's the second question:

Answer 4

Both questions are asking to evaluate for the definite integral for the integral at the top of the page.

Answer 5

What are the bounds for both functions?

Answer 6

I can't read them.

Answer 7

There is no bounds... I'm looking for the indefinite integral.

Answer 8

Oh for number 12 i mean... the bounds for number 10 is 8 to 15.

Answer 9

I meant the little numbers, at the top and bottom of the integral sign?

Answer 10

bottom = 8
top = 15

for the first question I posted.

Answer 11

Then, for that one I got: .8803

Answer 12

Second one is?

Answer 13

actually doesn't matter

Answer 14

there is no bounds, it's an indefinite integral.

Answer 15

Second one I got:

\[\int\limits \sin 2x \sin 6x \ dx = \frac{ 1 }{ 2 }\sin^3 (2x) \cos (2x)+C\]

Answer 16

Alright...but can you please explain to me how and where i went wrong?

Answer 17

Oops, can't right now.. I have to go somewhere with my GF. @Hero, mind explaining this?. I GTG.

Answer 18

kay. thanks anyways...

Answer 19

is there anyone out there that can please explain what i did wrong?

Answer 20

That has to be correct. Wolfram Alpha is never wrong

Answer 21

Here's the complete steps to find the integral

Answer 22

Ok! Thank you!
How did you get the full step by step from Wolfram Alpha?