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.
For problem 2, the integral from 0 to 1 of ((x^2)+1)(e^-x) dx, you are on the right track with your choices for u and dv. u = e^-x and dv = (x^2 +1) dx.
To solve this integral using the uv-integral v du formula, you need to find du and v. Differentiating u = e^-x gives you du = -e^-x dx. Integrating dv = (x^2 +1) dx gives you v = (1/3) x^3 + x.
Now, you can apply the formula:
∫((x^2)+1)(e^-x) dx = (uv - ∫v du)
Substituting the values you found:
= [(1/3) x^3 + x)(e^-x) - ∫((1/3) x^3 + x)(-e^-x) dx]
Simplifying this expression will give you the answer to the integral.
For problem 3, the integral of (x^5)cos(x^3)dx, you made a mistake with the substitution.
Let's try a different approach. Let u = x^3, then du = 3x^2 dx.
Substituting these into the integral, you get:
∫(x^5)cos(x^3)dx = (1/3)∫u cos(u) du
Now, you can integrate (1/3)u cos(u) using the uv-integral v du formula.
Choose v = sin(u) and du = 1 du. Integrating v du gives you uv = - 1/3 cos(u).
Applying the formula:
∫(x^5)cos(x^3)dx = (1/3)(-cos(u))(u) - ∫(-cos(u))(1) du
Simplifying this expression will give you the answer to the integral.