Newton Leibnitz's Theorem

Let $a, b, c$ be non-zero real numbers such that ; $\underset{0}{\overset{1}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx=\underset{0}{\overset{2}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx$, then the quadratic equation $a{x}^2+bx+c=0$ has -

If $x={\int }_0^{t^2}{e}^{\sqrt{z}}\left\{\frac{2\tan \sqrt{z}+1-{\tan }^2\sqrt{z}}{2\sqrt{z}{\sec }^2\sqrt{z}}\right\}dz$ & $y={\int }_0^{t^2}{e}^{\sqrt{z}}\left\{\frac{1-{\tan }^2\sqrt{z}-2\tan \sqrt{z}}{2\sqrt{z}{\sec }^2\sqrt{z}}\right\}dz$.

Then the inclination of the tangent to the curve at $t=\frac{\pi }{4}$ is-

Let $y=f\left(x\right)$ be a differentiable curve satisfying ${\int }_2^{x}f\left(t\right)dt=\frac{x^2}{2}+{\int }_x^2{t}^{2}f\left(t\right)dt$, then ${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\left\{\frac{f\left(x\right)+x^9-x^3+x+1}{{\cos }^{2}x}\right\}dx$ equals-

The value of $\underset{x\to \infty }{\lim }\left[\frac{d}{dx}\left\{{\int }_{\sqrt{3}}^{\sqrt{x}}\left(\frac{r^3}{\left(r+1\right)\left(r-1\right)}\right)dr\right\}\right]$ is-

Given a function $f\left(x\right)$ such that

$\left(a\right)$ it is integrable over every interval on the real line and

$\left(b\right)$ $f\left(T+x\right)=f\left(x\right)$, for every $x$ and a real $T$, then show that the integral ${\int }_a^{a+T}f\left(x\right)dx$ is independent of $a$.

The value of$\underset{x\to +\infty }{\lim }\frac{{\left({\int }_0^x{e}^{x^2}dx\right)}^2}{{\int }_0^x{e}^{2x^2}dx}$ is equal to

The value of $\underset{x\to 0^{+}}{\lim }\left\{\frac{{\int }_0^{x^2}\sin \sqrt{x}dx}{x^3}\right\}$ is equal to $A$, then $6A$ is equal to

If $f\left(x\right)={\int }_{e^x}^{e^{3x}}\left(\frac{t}{\ln t}\right)dt, x>0$, then differential coefficient of $f\left(x\right)$ w.r.t. $\ln x$, when $x=\ln 2$, is equal to

Investigate for maxima & minima for the function, $f\left(x\right)={\int }_1^x\left[2\left(t-1\right){\left(t-2\right)}^3+3{\left(t-1\right)}^2{\left(t-2\right)}^2\right]dt$.

Number of values of $x$ satisfying the equation ${\int }_{-1}^x\left(8t^2+\frac{28}{3}t+4\right)dt=\frac{\left(\frac{3}{2}\right)x+1}{{\log }_{\left(x+1\right)}\sqrt{x+1}}$, is

The function $f\left(x\right)={\int }_0^x\sqrt{1-t^4}dt$ is such that :

Let $f\left(x\right)={\int }_2^x\frac{dt}{\sqrt{1+t^4}}$ and $g$ be the inverse of $f.$ Then the value of $g^{\text{'}}\left(0\right)$ is

Let $f:R\to R$ be a differentiable function and $f\left(1\right)=4$.

Then, the value of $\underset{x\to 1}{\lim }{\int }_4^{f\left(x\right)}\frac{2t}{x-1}dt$ is

If $a, b$ and $c$ are real numbers, then the value of $\underset{t\to 0}{\lim }\ln \left(\frac{1}{t}{\int }_0^t{\left(1+a\sin bx\right)}^{\frac{c}{x}}dx\right)$ equals