Posted by **Shelley** on Tuesday, January 23, 2007 at 2:46pm.

Hello,

I have some calculus homework that I can't seem to get started..at least not on the right track? I have 3 questions

1. integral of [(p^5)*(lnp)dp]

I'm using the uv-integral v du formula

So first, I'm finding u and I think it's lnp.......so du is 1/p

Then, I think dv = p^5 and v is 1/6 p^6

But then I get (1/6)p^6 lnp - 1/42 p^7

Is that on the right track?

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2. integral from 0 to 1 of ((x^2)+1)(e^-x) dx

is u = e^-x and du equal -e^-x and

dv = x^2 +1 and v = 1/3 x^3 +x

That's as far as I got there..

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And 3. integral of (x^5)cos(x^3)dx

so my u = x^3 and du = 3x^2

so (1/3)x^7 * du = x^5?

Then I have integral of (1/3)x^7 * du *cos (u)

Then sin(x^3) is the integral?

___________ Thank you for your help!

But then I get (1/6)p^6 lnp - 1/42 p^7

You should get

(1/6)p^6 lnp - 1/36 p^6

You forgot to multiply by the derivative of ln(p) which is 1/p before integrating that term.

Another way to do this integral is by using the fact that:

p^x = exp[x log(p)]

This means that the derivative of p^x w.r.t. x is p^x log(p)

To calculate the integral, you just evaluate the integral of

p^(5+x)dp which is:

p^(6+x)/[6+x]

Now differentiate both sides w.r.t. x at x = 0.

ok, I think I understand...thank you!

2. integral from 0 to 1 of ((x^2)+1)(e^-x) dx

Do a partial integration, just like you did in problem 1.

Alternatively, you can first calculate the integral from zero to 1 of:

exp[-ax]dx

which is:

[1-exp(-a)]/a

for a = 1 you obtain one term of the integration, the integral of exp(-x). To obtain the integral of x^2 exp(-x) you differentiate the above integral twice w.r.t. a. That will bring down a factor x^2 inside the integral. Then you set a equal to 1.

And 3. integral of (x^5)cos(x^3)dx

You made a mistake with the substitution.

If you put u = x^3, then you can calculate du = 3 x^2 dx to see if you can get rid of the other factors easily. But in general, you should first solve for x:

x = u^(1/3)

and the calculate dx:

dx = 1/3 u^(-2/3)

If you insert for x the function u^(1/3)

and replace dx by 1/3 u^(-2/3), you see that the integral is:

1/3 u cos(u) du.

Youi can calculate this by partial integration just like the previous two problems. Or you can again use the differentiation w.r.t. a well chosen parameter trick. In this case you can do that by calculating the integral of:

1/3 sin(a u) du

which is:

-1/3 cos(a u)/a

Differentiation of both sides w.r.t. the parameter a at a = 1 gives you the result.