The value of integral $\int_0^1 {{e^{{x^2}}}} dx$ lies in interval

If $f(x)$ is a quadratic in $x$ , then $\int\limits_0^1 {f(x) dx}$ is

Let $\mathrm{a}$ and $\mathrm{b}$ be real constants such that the function $f$ defined by $f(x)=\left\{\begin{array}{cc}x^2+3 x+a & x \leq 1 \\ b x+2, & x>1\end{array}\right.$ be differentiable on $R$. Then, the value of $\int_{-2}^2 f(x) d x$ equals

The number of continuous functions $f :\left[0, \frac{3}{2}\right] \rightarrow(0, \infty)$ satisfying the equation $4 \int \limits_0^{3 / 2} f(x) d x+125 \int \limits_0^{3 / 2} \frac{d x}{\sqrt{f(x)+x^2}}=108$ is

$\int\limits_{ - 1}^{\frac{3}{2}} {|x\sin \pi x|dx} $ equals

Let $y=f(x)$ be a thrice differentiable function in $(-5,5)$. Let the tangents to the curve $y=f(x)$ at $(1, \mathrm{f}(1))$ and $(3, \mathrm{f}(3))$ make angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$, respectively with positive $x$-axis. If $27 \int_1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3} \quad$ where $\alpha, \quad \beta$ are integers, then the value of $\alpha+\beta$ equals