# Help in integration

• May 18th 2009, 08:58 PM
kn336a
Help in integration
Hi, I could not figure out two questions in my homework. The first problem deals with a piecewise function.

9
http://upload.wikimedia.org/math/f/e...eb485cd0f6.pngg(x)dx, where g(x)={1, x> 0 and x^3, 1 < x < 4 and sqrt x, 4 < x < 9
0

also,

Find the area below the interval [-2, -1], but above the curve y=x^3

Does this just mean
-1
-2

??
• May 18th 2009, 09:35 PM
pickslides
Quote:

Originally Posted by kn336a
Hi, I could not figure out two questions in my homework. The first problem deals with a piecewise function.

9
http://upload.wikimedia.org/math/f/e...eb485cd0f6.pngg(x)dx, where g(x)={1, x> 0 and x^3, 1 < x < 4 and sqrt x, 4 < x < 9
0

also,

Find the area below the interval [-2, -1], but above the curve y=x^3

Does this just mean
-1
-2

??

For Q1 try

$I_n=\int_0^1 1 dx + \int_1^4 x^3 dx + \int_4^9 \sqrt{x} dx$

For Q2 you are correct.
• May 18th 2009, 10:06 PM
Amer
Ok hi

I attach my answer the second question you just should multiply the answer by -1

Attachment 11481

Attachment 11482
• May 18th 2009, 10:07 PM
Amer
I was late sorry (Surprised)
• May 18th 2009, 10:26 PM
kn336a
Thanks guys! I got 929/12 for the first one or roughly 77.4167

the second one I got -15/4

is this right? I hope so!
• May 18th 2009, 10:42 PM
Amer
you answer is correct but the second I think you should take the absolute value cuz it is area and the area positive value

(Talking)
• May 18th 2009, 11:00 PM
kn336a
sorry, I'm a little confused. So do I just add an absolute value sign after the integration sign, then calculate?

I've gotten negative numbers for some of my integration problems and didn't get marked wrong. Is it because it says "but above the curve y=x^3?"

I thought that just meant the area between the interval and the curve. Thanks!