Integration problem

**ixi** · Jun 1st 2010, 11:56 AM

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

**skeeter** · Jun 1st 2010, 12:07 PM

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 $-3u^{-3}$ ... what do you get?

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

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

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

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

$ k = 3 $

**AllanCuz** · Jun 1st 2010, 12:07 PM

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

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

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

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

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

$k = 3$

**AllanCuz** · Jun 1st 2010, 12:08 PM

Originally Posted by skeeter

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

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

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

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

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

$ k = 24 $

The integral is from 0 to b, not 1 to b!

**skeeter** · Jun 1st 2010, 12:21 PM

Originally Posted by AllanCuz

The integral is from 0 to b, not 1 to b!

thank you

Thread: Integration problem

LinkBack

Thread Tools

Display

Integration problem

Similar Math Help Forum Discussions

integration problem

Simple Integration problem: Trigonometric Integration

Integration Problem

Integration problem !!

Integration problem

Search Tags