# Integration problem

• Jun 1st 2010, 11:56 AM
ixi
Integration problem
It's the integration of k(1+x)^-4 from 0 to infinity which is equal to 1. Solve for k.

My solution is ---
Let u=1+x du=dx
Then, it's k(-3(1+x)^-3) from 0 to infinity, which means k=1/3

But the solution manual says it's
(-k/3)(1/(1+3)^3) from 0 to infinity giving k=3

I'm not sure why it's -k/3 and not -3k

I thought when integration u^-4 it becomes -3u^-3
• Jun 1st 2010, 12:07 PM
skeeter
Quote:

Originally Posted by ixi
It's the integration of k(1+x)^-4 from 0 to infinity which is equal to 1. Solve for k.

My solution is ---
Let u=1+x du=dx
Then, it's k(-3(1+x)^-3) from 0 to infinity, which means k=1/3

But the solution manual says it's
(-k/3)(1/(1+3)^3) from 0 to infinity giving k=3

I'm not sure why it's -k/3 and not -3k

I thought when integration u^-4 it becomes -3u^-3

take the derivative of $\displaystyle -3u^{-3}$ ... what do you get?

$\displaystyle \lim_{b \to \infty} \int_0^b \frac{k}{(x+1)^4} \, dx = 1$

$\displaystyle \lim_{b \to \infty} \left[-\frac{1}{3}\frac{k}{(x+1)^3} \right]_0^b = 1$

$\displaystyle -\frac{k}{3} \lim_{b \to \infty} \left[\frac{1}{(b+1)^3} - \frac{1}{(0+1)^3} \right] = 1$

$\displaystyle \frac{k}{3} \cdot \frac{1}{1} = 1$

$\displaystyle k = 3$
• Jun 1st 2010, 12:07 PM
AllanCuz
Quote:

Originally Posted by ixi
It's the integration of k(1+x)^-4 from 0 to infinity which is equal to 1. Solve for k.

My solution is ---
Let u=1+x du=dx
Then, it's k(-3(1+x)^-3) from 0 to infinity, which means k=1/3

But the solution manual says it's
(-k/3)(1/(1+3)^3) from 0 to infinity giving k=3

I'm not sure why it's -k/3 and not -3k

I thought when integration u^-4 it becomes -3u^-3

$\displaystyle k \int_0^{ \infty } (1+x)^{-4} dx = 1$

$\displaystyle - \frac{k}{3} (1+x)^{-3} |_0^{ \infty } = 1$

$\displaystyle - \frac{k}{3} \lim_{ r \to \infty } (1+x)^{-3} |_0^{ R } = 1$

$\displaystyle - \frac{k}{3} ( 0 - 1 ) = 1$

$\displaystyle k = 3$
• Jun 1st 2010, 12:08 PM
AllanCuz
Quote:

Originally Posted by skeeter
take the derivative of $\displaystyle -3u^{-3}$ ... what do you get?

$\displaystyle \lim_{b \to \infty} \int_0^b \frac{k}{(x+1)^4} \, dx = 1$

$\displaystyle \lim_{b \to \infty} \left[-\frac{1}{3}\frac{k}{(x+1)^3} \right]_0^b = 1$

$\displaystyle -\frac{k}{3} \lim_{b \to \infty} \left[\frac{1}{(b+1)^3} - \frac{1}{(1+1)^3} \right] = 1$

$\displaystyle \frac{k}{3} \cdot \frac{1}{8} = 1$

$\displaystyle k = 24$

The integral is from 0 to b, not 1 to b!
• Jun 1st 2010, 12:21 PM
skeeter
Quote:

Originally Posted by AllanCuz
The integral is from 0 to b, not 1 to b!

thank you