Let $f(x) = \min \{[x-1], [x-2], \ldots, [x-10]\}$ where $[t]$ denotes the greatest integer $\leq t$. Then $\int_{0}^{10} f(x) \, dx + \int_{0}^{10} (f(x))^2 \, dx + \int_{0}^{10} |f(x)| \, dx$ is equal to

The values of $\alpha$ which satisfy $\int_{\pi /2}^{\alpha} \sin x \, dx = \sin 2\alpha$,where $\alpha \in [0, 2\pi]$,are equal to:

Let $f(x) = \int\limits_0^x {(t^2 + 2t + 2)dt}$ where $x$ is the set of real numbers satisfying the inequation $\log_{\sqrt{2}}(1 + \sqrt{6x - x^2 - 8}) \ge 0$. If the range of $f(x)$ is $[a, b]$,then $(a + b)$ is:

Let $f$ be a real-valued continuous function on $[0, 1]$ and $f(x) = x + \int_{0}^{1} (x - t) f(t) dt$. Then which of the following points $(x, y)$ lies on the curve $y = f(x)$?

Let $f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$ be a continuous function such that $f(0)=1$ and $\int_0^{\frac{\pi}{3}} f(t) dt = 0$. Then which of the following statements is (are) $TRUE$?
$(A)$ The equation $f(x) - 3 \cos 3x = 0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
$(B)$ The equation $f(x) - 3 \sin 3x = -\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
$(C)$ $\lim_{x \rightarrow 0} \frac{x \int_0^x f(t) dt}{1 - e^{x^2}} = -1$
$(D)$ $\lim_{x \rightarrow 0} \frac{\sin x \int_0^x f(t) dt}{x^2} = -1$

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