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

Answer 1

Emma Aldridge

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

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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Answer 2

Ezekiel Alexander

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.