Solve for a.
1 answer:
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First we find the zeroes so we don't take the integral of negative bits
4x-x²
x(4-x)
zeroes at x=0 and x=4
it opens down
so the part we are interested in is the bit between x=0 and x=4
![\int\limits^4_0 {4x-x^2} \, dx =[2x^2- \frac{1}{3}x^3]^4_0=(32- \frac{64}{3})-(0)= 10.6666666666](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E4_0%20%7B4x-x%5E2%7D%20%5C%2C%20dx%20%3D%5B2x%5E2-%20%5Cfrac%7B1%7D%7B3%7Dx%5E3%5D%5E4_0%3D%2832-%20%5Cfrac%7B64%7D%7B3%7D%29-%280%29%3D%20%2010.6666666666)
or aout 10 and 2/3
C is answer
4x-x²
x(4-x)
zeroes at x=0 and x=4
it opens down
so the part we are interested in is the bit between x=0 and x=4
C is answer
