The value of the integral $\int_0^1 e^{x^2} dx$ lies in the interval

$f(x) = \begin{cases} x^2, & 0 \leq x < 1 \\ \sqrt{x}, & 1 \leq x \leq 2 \end{cases} \implies \int_0^2 f(x) \, dx = ?$

$\int_0^a \frac{x-a}{x+a} dx =$

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

Let $f(x) = \begin{cases} \frac{1}{3}, & x \le \pi/2 \\ \frac{b(1-\sin x)}{(\pi-2x)^2}, & x > \pi/2 \end{cases}$. If $f$ is continuous at $x = \pi/2$, then the value of $\int_0^{3b-6} |x^2+2x-3| dx$ is:

The value of $\int_1^5 (|x - 3| + |1 - x|) \, dx$ is