Hi can you please explain this to me,

For

$\displaystyle \int^k_0 x(x-4)\ dx = 13$, 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

$\displaystyle \int^k_0 x^3(x-9)\ dx = 13$

Made this one up myself, I just wanted to know how to get k=4 in the previous question..

Please help, thank you so much!