Integration by parts question?

Question

I'm having some trouble figuring this one out.

Integral 0 to 1

r^3/(sq root(4+r^2)) dr

Would someone mind helping me out :D?

Thanks

Answer 1

Integral (0 to 1, r^3 / sqrt(4 + r^2) ) dr

I'm not sure you can solve this one using integration by parts. You have to use substitution.

First things first, let's split up r^3 into r^2 and r. This will give us

Integral (0 to 1, [r^2 / sqrt(4 + r^2)] r dr )

Notice I moved the r next to the dr to prove a point.

Let u = 4 + r^2. Then
u - 4 = r^2
du = 2r dr. Multiplying both sides by (1/2), we get
(1/2) du = r dr

Note that (1/2) du is equal to r dr, which is the tail end of our integral. Now, we substitute accordingly, and get

Integral ( [ (u - 4) / sqrt(u)] (1/2) du )

Pulling out the constant (1/2) out of the integral, we get

(1/2) Integral ( [ (u - 4) / sqrt(u) ] du )

We can split that fraction up into two fractions. Note that sqrt(u) is equal to u^(1/2), so let's change that right away.

(1/2) Integral ( u/u^(1/2) - 4/u^(1/2) ) du

(1/2) Integral ( u^(1/2) - 4u^(-1/2) ) du

Now, we can integrate this normally, using the reverse power rule.

(1/2) [ (2/3)u^(3/2) - 4 (2u^(1/2)) ]

(1/2) [ (2/3)u^(3/2) - 8u^(1/2) ]

Notice that this is a definite integral, with our bounds formerly from 0 to 1. Now that we used substitution, our bounds may change. Since we let u = r^2, then
If r = 0, then u = 0^2 = 0
If r = 1, then u = 1^2 = 1
By coincidence only, our bounds did not change. That's what we want to evaluate this integral on:

(1/2) [ (2/3)u^(3/2) - 8u^(1/2) ] {evaluated from 0 to 1}

(1/2) { [ (2/3)(1) - 8(1) ] - [0 - 0] }
(1/2) { 2/3 - 8 }
(1/2) { 2/3 - 24/3 }
(1/2) [-22/3} = -11/3

Answer 2

The trick is to come up with good choices of u and dv to fit the equation integral u * dv = u*v - integral v * du

The best choice took me a second to see, but it turns out to be
u = r^2
dv = r / sqrt(4+r^2)*dv

If you integrate to find v, it is simply:
v = sqrt(4+r^2)
du = 2r*dr, and so...

original integral = r^2 * sqrt(4+r^2) - integral(2r * sqrt(4 + r^2))dr

You can do the second integral by u substiution

integral(2r * sqrt(4 + r^2))dr
let u = 4+r^2
du = 2r

integral(2r * sqrt(4 + r^2))dr = integral(sqrt(u))du = (2/3)*u^(3/2)
= (2/3)(4+r^2)^(3/2)

Therefore, the original integral is just:

r^2 * sqrt(4+r^2) - (2/3)(4+r^2)^(3/2) + C (don't forget the constant!)

Because you want to definite integral, we can just plug in the upper and lower bounds (0 to 1) and get:

(1^2 * sqrt(4+1^2) - (2/3)(4+1^2)^(3/2)) - (0^2 * sqrt(4+0^2) - (2/3)(4+0^2)^(3/2)) = sqrt(5) - (2/3)5^(3/2) + (2/3)4^(3/2) = 0.115

Answer 3

u^2 = 4+r^2, u>0
u du = r dr
∫r^3/(sq root(4+r^2)) dr
= ∫(u^2-4)du
= (1/3)u^3 - 4u

r=0, u=2
r=1, u=sqrt(5)

∫r^3/(sq root(4+r^2)) dr from 0 to 1
= ∫(u^2-4)du from 2 to sqrt(5)
= 0.1158