# I need help!

• Jul 29th 2006, 04:57 AM
margaritas
I need help!
Hey guys, lol I'm here again! with an integration question. I'm having a quiz on the topic this Monday and while doing revision, I couldn't quite get the following question:

Evaluate (integrate; upper limit - 2, lower limit - 0) absolute (x^2 + 2x - 3) dx.

I got 2/3 but the answer at the back says it's 4?
• Jul 29th 2006, 05:19 AM
earboth
Quote:

Originally Posted by margaritas
Hey guys, lol I'm here again! with an integration question. I'm having a quiz on the topic this Monday and while doing revision, I couldn't quite get the following question:
Evaluate (integrate; upper limit - 2, lower limit - 0) absolute (x^2 + 2x - 3) dx.
I got 2/3 but the answer at the back says it's 4?

Hello, margaritas,

I've attached a diagram to show you what you have to calculate.

1. Your function has a zero at x = 1. So you have to split the integral into:

$\int_{0}^{1}(-(x^2+2x-3))dx=\left[-\frac{1}{3}x^3-x^2+3x\right]_{0}^{1}=\frac{5}{3}$ and
$\int_{1}^{2}(x^2+2x-3)dx=\left[\frac{1}{3}x^3+x^2-3x\right]_{1}^{2}=\frac{7}{3}$

Add these partial values and you'll get 4.

Greetings

EB
• Jul 29th 2006, 05:21 AM
margaritas
That's fast, thanks for your help once again, EB! :)

I'll take a look at your solution straight after I finish a report for a project. Yes I have loads of homework I know. lol