# Area under curve

• Jan 13th 2013, 06:04 AM
Tutu
Area under curve
Hi can you please explain this to me,

For
$\int^k_0 x(x-4)\ dx = 13$, why must k be from 0 to 4 such that the area is 13units^2?
I integrated x(x-4) from 0 to 4, I only got the area as 10.something...it's not 13?

If that is so, then what should the value for k be such that
$\int^k_0 x^3(x-9)\ dx = 13$
Made this one up myself, I just wanted to know how to get k=4 in the previous question..

• Jan 13th 2013, 06:30 AM
Re: Area under curve
Yep you're correct if you integrate that you'll get F(x)=(1/3)x^3-2x^2+c and when you integrate that from 0 to 4 you get 10.67. Might be a typo.
• Jan 13th 2013, 08:00 AM
christophina
Re: Area under curve
Plot the graph of f(x)=x(x-4). It's a parabola with roots 0 and 4.

Attachment 26555

If you use limits from 0 to 4, you'll get -32/3 which indicates an area BELOW the x-axis. So k has to be greater than 4, if you want the value of the definite integral to be 13 for your answer.
• Jan 14th 2013, 05:08 AM
Tutu
Re: Area under curve
Thank you I understand!