Question

Let $F\left(x\right)={\int }_0^{x^2}{t}^2-5t+4dt$

Answer 1

$F\left(x\right)={\int }_0^{x^2}(t^2-5t+4)dt$

On Leibnitz Formula:-

$F^{\prime }\left(x\right)=\frac{d}{dx}{\int }_0^{x^2}(t^2-5t+4)dt=(x^2-5x^2+4)2x-0$

For monotonicity, first we find $F^{\prime }\left(x\right)=0$

$\therefore (x^4-5x^2+4)2x=0$

$2x(x^2-4)(x^2-3)=0$

$x(x-2)(x+2)(x-1)(x+1)=0$

Refer image.

$F\left(x\right)$ increases in interval $(-2,-1)\cup (0,1)\cup (2,\infty )$

$F\left(x\right)$ decreases in interval $(-\infty ,-2)\cup (-1,0)\cup (1,2)$

Now,

$F\left(1\right)={\int }_0^1{t}^2-5t+4dt={[\frac{t^3}{3}-5\frac{t^2}{2}+4t]}_0^1=\frac{1}{3}-\frac{5}{2}+4$

$=\frac{11}{6}$

$F\left(\sqrt{2}\right)={\int }_0^2{t}^2-5t+4dt={[\frac{t^3}{3}-\frac{5t^2}{2}+4t]}_0^2$

$=\frac{8}{3}-\frac{20}{2}+8=\frac{2}{3}$