need calculus 2 help - definite integrals?

Question

problem is: compute the definte integral

from 0 to ln2

8e^(7x) sin(pi(e^(7x))dx

answer choices are:

1. -112
2. -(16/7)(1/pi)
3. -(16/pi)
4. -(16/7)

TIA

Answer 1

Answer choice (2) is correct. Here's why:

Make the substitution y = e^(7x).
Then dy = 7e^(7x) dx = 7y dx
and dx = 1/7 (dy/y).

For limits, when x=0, y=1
and when x = ln 2, y = e^(7 ln 2) = e^[ln(2^7)] = 2^7

Now we integrate 8y sin(pi y) (1/7) (dy/y) = (8/7) sin(pi y) dy from 1 to 2^7.

That integral is -8/(7 pi) cos(pi y) which we evaluate from 1 to 2^7.

Upper limit first. Since 2^7 is an even integer, cos(2^7 * pi) = 1.
Now the bottom limit. For y=1, cos pi = -1.

Doing the entire expression, we have

-8/(7 pi)[1 - (-1)] = -16/(7 pi)

So Answer (2) is right.

Answer 2

Ok, let's see, this word pad does not allow to show the symbols, so I will do my best to explain the concepts as we go...

INTEGRATION[8e^(7x) sin(pi(e^(7x))) dx] over the prescribed limits

let u = pi(e^(7x))
du = 7pi e^(7x) dx

substitute in the original equation and you will get

INTEGRAL [8/(7pi) sin(u) du] different set of limits

The integration is simple (integral of a sine(x) is a -cosine(x))

so after some manipulation you will get,

-8/(7pi) * (1+1)

or -16/(7*pi) or answer 2

There you go!

Answer 3

Sounds like you need a little help on your homework. You should notice the exponential (e^7x) both inside the sine argument and outside. You also know the special property of e^z regarding integrals and differentials. This should suggest to you some sort of substitution that converts the integral to sin(u) du. That should be enough to get you going.

Answer 4

put pie^7x=t and integrate and then apply the limits