# Calculus

posted by .

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?

***********************************
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..
********************************
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.

## Similar Questions

1. ### integration by part

s- integral s (p^5) lnp dp ? = (lnp)[(p^6)/(6)] - s (i'm confused here) Let ln p = u and p^5 dv = dv du = 1/p and v = p^6/6 The integral can be written uv - v du = (ln p*p^6)/6 - S (p^5/6) = (1/6)ln p*p^6 - ?
2. ### Calculus

Hello, I'm having trouble with this exercise. Can you help me?
3. ### Calculus III

I'm supposed to find the expression 'y', when (x^2+100)y'=23 When y(10sqrt(3))=0 Also, I have to find the indefite integral of xsqrt(19x-7), using integration tables in the back of the book "Calculus of a Single Variable, 8th Edition." …
4. ### Calculus - Integrals

I have 3 questions, and I cannot find method that actually solves them. 1) Integral [(4s+4)/([s^2+1]*([S-1]^3))] 2) Integral [ 2*sqrt[(1+cosx)/2]] 3) Integral [ 20*(sec(x))^4 Thanks in advance.
5. ### Calculus - Integrals

I have 3 questions, and I cannot find method that actually solves them. 1) Integral [(4s+4)/([s^2+1]*([S-1]^3))] 2) Integral [ 2*sqrt[(1+cosx)/2]] 3) Integral [ 20*(sec(x))^4 Thanks in advance.
6. ### calculus

A question on my math homework that I can't seem to solve... Rotate the region bounded by y=x^2-3x and the x-axis about the line x=4. Set up the integral to find the volume of the solid. I'm pretty sure that the integral is in terms …
7. ### Calculus

I have two questions, because I'm preparing for a math test on monday. 1. Use the fundamental theorem of calculus to find the derivative: (d/dt) the integral over [0, cos t] of (3/5-(u^2))du I have a feeling I will be able to find …
8. ### Calculus

Hi there i am having some problems trying to do my calculus homework and i really need help on how to show the step to proof the volume of a sphere which is V= 4/3pirsquare. But I have to use triple integral to proof the volume of …
9. ### Calc

Evaluate the integral using any method: (Integral)sec^3x/tanx dx I started it out and got secx(1tan^2x)/tanx. I know I just have to continue simplifying and finding the integral, but I'm stuck on the next couple of steps. Also, I have …
10. ### Calculus II

Integrate using integration by parts (integral) (5-x) e^3x u = 5-x du = -dx dv = e^3x v = 3e^3x I wonder if this is right so far. = uv - (integral) v du = (5-x)(3e^3x) - (integral) (-3e^3x) =(5-x)(3e^3x) + (integral) (3e^3x) = (5-x)(3e^3x) …

More Similar Questions