Question

The point of inflection of the function $y=\underset{0}{\overset{x}{\int }}(t^2-3t+2)dt$ is :

• A. $(\frac{1}{2},\frac{3}{2})$
• B. $(\frac{3}{2},\frac{3}{4})$
• C. $(\frac{-3}{2},-\frac{3}{4})$
• D. $(\frac{-1}{2},-\frac{3}{2})$

Answer 1

The correct option is B $(\frac{3}{2},\frac{3}{4})$

Using Newton -Leibnitz's rule
$\frac{dy}{dx}=x^2-3x+2$

For inflection point,
$\frac{d^{2}y}{d{x}^2}=2x-3=0\Rightarrow x=\frac{3}{2}\Rightarrow y={[\frac{x^3}{3}-\frac{3x^2}{2}+2x]}_0^{3/2}=\frac{3}{4}$