Question

The total number of extremum(s) of $y={\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}dt$ are __________

Answer 1

Given : $y={\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}dt$

Using Leibniz integral rule
Differentiate the equation $y$ with respect to $x$, we get

$y^{\prime }=2x\frac{x^4-5x^2+4}{2+e^{x^2}}$

Now, $y^{\prime }=0$

$\Rightarrow 2x\frac{x^4-5x^2+4}{2+e^{x^2}}=0$
Taking the denominator to right which make it zero
$\Rightarrow 2x(x^4-5x^2+4)=0$

$\Rightarrow 2x(x^2-4)(x^2-1)=0$
Using maxima and minima
$x=0,x=2,x=-2,x=1,x=-1$
Hence, the total number of extremum is $5$.