If $\int_0^{2a} x^2 \sqrt{2ax-x^2} dx = ka^4$, then $k : \pi =$ (in $:8$)

The value of $f(1)$,given the equation $\int_0^{x^2} x f(t) dt = x^5 - x^3$ is

The points of extremum of $\int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$ are

Let $f : R \rightarrow R$ be a continuous odd function,which vanishes exactly at one point and $f(1) = \frac{1}{2}$. Suppose that $F(x) = \int_{-1}^x f(t) dt$ for all $x \in [-1, 2]$ and $G(x) = \int_{-1}^x t|f(f(t))| dt$ for all $x \in [-1, 2]$. If $\lim_{x \rightarrow 1} \frac{F(x)}{G(x)} = \frac{1}{14}$,then the value of $f\left(\frac{1}{2}\right)$ is

$\int_0^{\pi/6} \cos^4 3\theta \cdot \sin^2 6\theta \, d\theta$ is equal to

$\int_{-\pi}^\pi \frac{\cos ^{2022} x}{1+(2022)^x} d x=$