Question

If $f\left(x\right)=\int \sqrt{(2x-x^2)}dx$ , then find the value of f(1)
[ Take the constant of integration equal to zero]

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Answer 1

Since there is an $x^2$ term and 2x, we will try to express it as a perfect square plus a constant. We also note that the sign of $x^2$ is negative. So, we will take -1 out from the bracket and proceed
We get $(2x-x^2$$=-(-2x+x^2)$
=$-(x^2-2x+1-1)$
=$-\left[\right(x-1{)}^2-1]$
= $1-(x-1{)}^2$
So we can write $\int \sqrt{(2x-x^2)}dx=\int \sqrt{1-(x-1{)}^2}dx$
This is of the form $\int \sqrt{(a^2-x^2)}dx$, which is equal to
$\frac{x}{2}\sqrt{(a^2-x^2)}+\frac{a^2}{2}si{n}^{-1}\left(x\right)+c$
In the integral $\int \sqrt{1-(x-1{)}^2}dx$ instead of ‘a’, we have 1 and instead of x, we have (x-1).Replacing, x and ‘a’ with, (x-1) and 1, we get f(x) equal to
$f\left(x\right)=\int \sqrt{1-(x-1{)}^2}dx=\frac{x-1}{2}\sqrt{1^2-(x-1{)}^2}+\frac{1}{2}si{n}^{-1}(x-1)$
We want to find f (1)
$\Rightarrow f\left(1\right)=0+0=0$
$\left[si{n}^{-1}\right(0)=0]$