Let $f$ be a twice differentiable function on $\mathbb{R}$. If $f^{\prime}(0)=4$ and $f(x)+\int_{0}^{x}(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x$,then $(2 a+1)^{5} a^{2}$ is equal to $\dots\dots$

Let $F(x) = \int_x^{x^2+\frac{\pi}{6}} 2 \cos^2 t \, dt$ for all $x \in \mathbb{R}$ and $f: [0, \frac{1}{2}] \rightarrow [0, \infty)$ be a continuous function. For $a \in [0, \frac{1}{2}]$,if $F'(a) + 2$ is the area of the region bounded by $x=0, y=0, y=f(x)$ and $x=a$,then $f(0)$ is

If $\int f(x) dx = F(x) + C$,then $\frac{d}{dt} \int_{g(t)}^{h(t)} f(x) dx =$

$\mathop {Lim}\limits_{k \to 0} \frac{1}{k} \int\limits_0^k (1 + \sin 2x)^{\frac{1}{x}} dx$

The value of $\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$ is equal to

Let $f(x) = \int\limits_0^{x^2} {(t - 1)(t - 4)(t - 9)} dt$,then: