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

First we solve the indefinite integral:

Now we see that:

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To find ( f(x) = \int x\sqrt{3-x} , dx ) when ( f(3) = 0 ), integrate ( x\sqrt{3-x} ) with respect to ( x ) and evaluate it at ( x = 3 ).

[ \begin{align*} f(x) &= \int x\sqrt{3-x} , dx \ &= -\frac{2}{3}(3-x)^{3/2} + C \end{align*} ]

Given ( f(3) = 0 ), we have:

[ 0 = -\frac{2}{3}(3-3)^{3/2} + C = -\frac{2}{3}(0)^{3/2} + C = 0 + C ]

Thus, ( C = 0 ).

Therefore, ( f(x) = -\frac{2}{3}(3-x)^{3/2} ).

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