The value of $\int_{0}^{1} 9x^8 dx + \int_{0}^{\pi/2} \cos x dx$ is

If $f(x)$ is a function satisfying $f^{\prime}(x)=f(x)$ with $f(0)=1$ and $g(x)$ is a function that satisfies $f(x)+g(x)=x^2$. Then the value of the integral $\int_0^1 f(x) g(x) d x$ is

Let $u = \int_{0}^{\infty} \frac{dx}{x^4 + 7x^2 + 1}$ and $v = \int_{0}^{\infty} \frac{x^2 dx}{x^4 + 7x^2 + 1}$. Then:

If $b_{n} = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{2} nx}{\sin x} dx$,$n \in N$,then

Assertion $(A)$: $\int_0^{\frac{\pi}{2}} (\sin^6 x + \cos^6 x) dx$ lies in the interval $(\frac{\pi}{8}, \frac{\pi}{2})$.
Reason $(R)$: $\sin^6 x + \cos^6 x$ is a periodic function with period $\frac{\pi}{2}$.

Let $f^{\prime}(x)=\frac{192 x^3}{2+\sin ^4 \pi x}$ for all $x \in R$ with $f\left(\frac{1}{2}\right)=0$. If $m \leq \int_{1 / 2}^1 f(x) d x \leq M$,then the possible values of $m$ and $M$ are