If $\int\limits_0^x {f\left( t \right)} dt = {x^2} + \int\limits_x^1 {{t^2}f\left( t \right)dt} $,then $f'(1/2)$ is

The value of $\mathop {\lim }\limits_{x \to 0} \left( \frac{\int_0^{x^2} \sec^2 t \, dt}{x \sin x} \right)$ is

The function $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$ is an increasing function in the interval ...

If $I=\int_{-a}^a(x^4-2x^2)dx$,then $I$ is minimum at $a=$

If $f(x) = \int_{\pi^2/16}^{x^2} \frac{\sin x \cdot \sin \sqrt{\theta}}{1 + \cos^2 \sqrt{\theta}} \, d\theta$,then the value of $f'(\frac{\pi}{2})$ is:

Let $f:(0, \infty) \rightarrow \mathbb{R}$ be given by $f(x)=\int_{\frac{1}{x}}^x e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t}$. Then
$(A)$ $f(x)$ is monotonically increasing on $[1, \infty)$
$(B)$ $f(x)$ is monotonically decreasing on $(0,1)$
$(C)$ $f(x)+f\left(\frac{1}{x}\right)=0$,for all $x \in(0, \infty)$
$(D)$ $f\left(2^x\right)$ is an odd function of $x$ on $\mathbb{R}$