# Integral with limit

• Feb 15th 2013, 12:33 PM
Steelers72
Integral with limit
Express the limit as a definite integral on the given interval.
 n xi ln(1 + xi2) Δx, [0, 4] http://www.webassign.net/wastatic/wa...ex/img/sum.gif i = 1

My work: I figured out that f(x) = xln(1+x^2)

therefore we need to find the integral from 0 to 4 of xln(1+x^2)dx

I'm a little confused. Do I start by making a graph? I haven't memorized the log graphs so sorry for my naivety. I appreciate any help.
• Feb 15th 2013, 01:16 PM
Plato
Re: Integral with limit
Quote:

Originally Posted by Steelers72
Express the limit as a definite integral on the given interval.
 n xi ln(1 + xi2) Δx, [0, 4] http://www.webassign.net/wastatic/wa...ex/img/sum.gif i = 1

My work: I figured out that f(x) = xln(1+x^2)

therefore we need to find the integral from 0 to 4 of xln(1+x^2)dx

Your work is correct. What more do you need to do?
It does not ask you to evaluate it. Does it?
• Feb 15th 2013, 02:29 PM
Prove It
Re: Integral with limit
Quote:

Originally Posted by Steelers72
Express the limit as a definite integral on the given interval.
 n xi ln(1 + xi2) Δx, [0, 4] http://www.webassign.net/wastatic/wa...ex/img/sum.gif i = 1

My work: I figured out that f(x) = xln(1+x^2)

therefore we need to find the integral from 0 to 4 of xln(1+x^2)dx

I'm a little confused. Do I start by making a graph? I haven't memorized the log graphs so sorry for my naivety. I appreciate any help.

If you need to evaluate the integral \displaystyle \displaystyle \begin{align*} \int_0^4{x\ln{\left( 1 + x^2 \right)} \, dx } = \frac{1}{2} \int_0^4{ 2x\ln{ \left( 1 + x^2 \right) } \,dx } \end{align*}, let \displaystyle \displaystyle \begin{align*} u = 1 + x^2 \implies du = 2x\, dx \end{align*} and note that \displaystyle \displaystyle \begin{align*} u(0) = 1 \end{align*} and \displaystyle \displaystyle \begin{align*} u(4) = 17 \end{align*}, then

\displaystyle \displaystyle \begin{align*} \frac{1}{2}\int_0^4{2x\ln{\left( 1 + x^2 \right) }\, dx} &= \frac{1}{2} \int_1^{17} { \ln{(u)} \, du } \end{align*}

which can now be solved by integration by parts.
• Feb 15th 2013, 08:03 PM
Steelers72
Re: Integral with limit
Thank you both; the work I had done was all we needed to do haha...sorry for that!