
Area under curve
Hi can you please explain this to me,
For
$\displaystyle \int^k_0 x(x4)\ dx = 13$, why must k be from 0 to 4 such that the area is 13units^2?
I integrated x(x4) 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
$\displaystyle \int^k_0 x^3(x9)\ dx = 13$
Made this one up myself, I just wanted to know how to get k=4 in the previous question..
Please help, thank you so much!

Re: Area under curve
Yep you're correct if you integrate that you'll get F(x)=(1/3)x^32x^2+c and when you integrate that from 0 to 4 you get 10.67. Might be a typo.

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Re: Area under curve
Plot the graph of f(x)=x(x4). 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 xaxis. So k has to be greater than 4, if you want the value of the definite integral to be 13 for your answer.

Re: Area under curve