If $I_1 = \int_0^{\pi / 4} \sin^2 x \, dx$ and $I_2 = \int_0^{\pi / 4} \cos^2 x \, dx$, then,

Let $f(x)$ be a differentiable function defined on $[0,2]$ such that $f^{\prime}(x) = f^{\prime}(2-x)$ for all $x \in (0,2)$,$f(0) = 1$ and $f(2) = e^{2}$. Then the value of $\int_{0}^{2} f(x) dx$ is ..... .

Let $f:[-2, 3] \to [0, \infty)$ be a continuous function such that $f(1-x) = f(x)$ for all $x \in [-2, 3]$. If $R_1$ is the numerical value of the area of the region bounded by $y = f(x)$,$x = -2$,$x = 3$ and the $x$-axis,and $R_2 = \int_{-2}^3 x f(x) dx$,then:

If $\int_0^{10} f(x) d x=5$,then $\sum_{k=1}^{10} \int_0^1 f(k-1+x) d x=$

The value of the integral $\int_{0}^{1} x \cot^{-1}(1 - x^2 + x^4) dx$ is

If $\int \limits_0^\pi \frac{5^{\cos x}(1+\cos x \cos 3x+\cos^2 x+\cos^3 x \cos 3x) dx}{1+5^{\cos x}} = \frac{k \pi}{16}$,then $k$ is equal to $...........$.