Question

The function $f$ whose graph passes through $(0,7/3)$ and whose derivative is $x\sqrt{1-x^2}$ is given by

• A. $f\left(x\right)=\left(1/3\right)\left(\right(1-x^2{)}^{3/2}+7)$
• B. $f\left(x\right)=\left(3/2\right)[{\sin }^{-1}x+6]$
• C. $f\left(x\right)=-\left(1/3\right)\left[\right(1-x^2{)}^{3/2}+\frac{8}{3}]$
• D. $f\left(x\right)=-\left(2/3\right)\left[\right(1-x^2{)}^{3/2}-8]$

Answer 1

The correct option is C $f\left(x\right)=-\left(1/3\right)\left[\right(1-x^2{)}^{3/2}+\frac{8}{3}]$

$f\left(x\right)=\int x\sqrt{1-x^2}dx$

Let $(1-x^2)=z⟹-2xdx=dz⟹dx=\frac{1}{2}dz$

$f\left(x\right)=\int \frac{-1}{2}\sqrt{z}dz$$=-\frac{1}{3}(1-x^2{)}^{3/2}+c$

On satisfying with $(0,7/3)$ we get $c=8/3$

$=-\frac{1}{3}(1-x^2{)}^{3/2}+8/3$