What is #f(x) = int xsqrt(3-x) dx# if #f(3) = 0 #?

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1 Answer
Andrea S.
Jan 3, 2017

#f(x) = -2/5(3-x)(x+2)sqrt(3-x)#

Explanation:

First we solve the indefinite integral:

#int xsqrt(3-x)dx = int (x-3+3) sqrt(3-x) dx = - int (3-x)^(3/2)dx +3 int(3-x)^(1/2)dx=2/5(3-x)^(5/2)-2(3-x)^(3/2)+C#

Now we see that:

#f(3) = C#, so #f(3) = 0 => C=0# and finally:

#f(x) = 2/5(3-x)^(5/2)-2(3-x)^(3/2)=2/5(3-x)sqrt(3-x)(3-x-5)=-2/5(3-x)(x+2)sqrt(3-x)#

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