The minimum value of the twice differentiable function $f(x) = \int_{0}^{x} e^{x-t} f'(t) dt - (x^2 - x + 1) e^x, x \in R$,is.

Let $f:(0, +\infty) \to \mathbb{R}$ and $F(x) = \int_0^{x^2} f(t) dt$. If $F(x) = x^2(1 + x)$,then $f(4)$ equals

If $f(x) = \int_{\pi^2/16}^{x^2} \frac{\sin x \cdot \sin \sqrt{\theta}}{1 + \cos^2 \sqrt{\theta}} \, d\theta$,then the value of $f'(\frac{\pi}{2})$ is:

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

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f(\frac{tx}{36}) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points $(2, 1)$ and $(-4, \beta)$, then $\beta^{\alpha}$ is equal to . . . . . . .

$\int_0^{\pi / 2} \sin ^m x \cos ^4 x \, dx = \frac{7 \pi}{2048} \Rightarrow m = ?$