$\int_0^1 {{e^{2\ln x}}dx} = $

The velocity of a particle which starts from rest is given by the following table. The total distance travelled (in metre) by the particle in $10 \ s$, using the Trapezoidal rule, is:

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

Evaluate $\int_{-2}^1 f(x) dx$, where $f(x) = \begin{cases} 1-2x, & x \leq 0 \\ 1+2x, & x \geq 0 \end{cases}$

Let $[t]$ denote the largest integer less than or equal to $t$. If $\int_0^3 \left( [x^2] + [\frac{x^2}{2}] \right) dx = a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}$,where $a, b, c \in \mathbb{Z}$,then $a + b + c$ is equal to:

$\int_{0}^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx =$

Let $(2^{1-a} + 2^{1+a})$, $f(a)$, $(3^a + 3^{-a})$ be in $A$.$P$. and $\alpha$ be the minimum value of $f(a)$. Then the value of the integral $\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}$ is: