Definite Integration

$\int_{0}^{\infty}\,e^{-x}dx$

If $\int_{0}^{1}\,f(x)dx=1$ and $\int_{1}^{2}\,f(y)dy=2$, then $\int_{0}^{2}\,f(z)dz=$

$y=ae^{mx}+be^{-mx}$ satisfies which of the following differential equations

Evaluate $\int_{-a}^{a} \,\{f(x)-f(-x)\}dx$

$\int_{1}^{3}\,\dfrac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}}dx$

The value of $\int_{2}^{4}(\{x\}+[x]) d x=\ldots$ (where $\{x\}$ and $[x]$ are the fractional part function and the greatest integer function, respectively)

$\int_{3}^{5} \dfrac{\sqrt{x} d x}{\sqrt{8-x}+\sqrt{x}}=$

$\int_{0}^{\pi/4}\,\dfrac{dx}{1+\cos 2x}$

$\int_{-1}^{1}\,x|x|dx$

$\int_{1}^{e^{2}}\,\dfrac{dx}{x(1+\log x)^{2}}=$

$\int_{0}^{\pi/2}\,\dfrac{\sin x\cos x}{1+\sin^{4}x}dx$

If $\int_{0}^{a} \dfrac{d x}{1+4 x^{2}}=\dfrac{\pi}{8}$, then $a=$

$\int_{0}^{1}\,\dfrac{3x^{3}-4x^{2}+1}{\sqrt{x}}dx$

If $f(x)=\left\{\begin{array}{l}1-2x,\,for\,x\leq 0 \\ 1+2x,\,for\,x\geq 0\end{array} \right)$, then $\int_{-1}^{1}\,f(x)dx=$

Let $\dfrac{d}{dx}(f(x))=\dfrac{e^{\sin x}}{x},x>0$. If $\int_{1}^{4}\,\dfrac{2e^{\sin x^{2}}}{x}dx=f(k)-f(1)$ then find one possible value of k .