$\left[ \int_{0}^{2} \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}} \, dx \right]$ is equal to (where $[\cdot]$ is the $G.I.F.$)

Suppose $g(x) = \int_0^x f(t) dt$,where $f$ is such that for $t \in [0, 1]$,$0 \le f(t) \le \frac{1}{2}$ and for $t \in [1, 2]$,$\frac{1}{2} \le f(t) \le 1$. Find the range of $g(2)$.

The value of $\int_{-1 / \sqrt{2}}^{1 / \sqrt{2}}\left(\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right)^{1 / 2} d x$ is:

Let $f(0)=1, f(0.5)=\frac{5}{4}, f(1)=2, f(1.5)=\frac{13}{4}$ and $f(2)=5$. Using Simpson's rule, $\int_0^2 f(x) dx$ is equal to

The value of the integral $\int_0^4 \frac{d x}{1+x^2}$ obtained by using the Trapezoidal rule with $h=1$ is

Evaluate the integral: $\int_0^\pi \left(\cos^2 \left(\frac{3\pi}{8} - \frac{x}{4}\right) - \cos^2 \left(\frac{11\pi}{8} + \frac{x}{4}\right)\right) dx$