If $$\mathrm{f}(x)$$ is a function satisfying $$\mathrm{f}^{\prime}(x)=\mathrm{f}(x)$$ with $$\mathrm{f}(0)=1$$ and $$\mathrm{g}(x)$$ is a function that satisfies $$\mathrm{f}(x)+\mathrm{g}(x)=x^2$$. Then the value of the integral $$\int_\limits0^1 f(x) g(x) d x$$ is

$$\int_\limits{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\mathrm{d} x}{1+\cos x}$$ is equal to

The value of the integral $$\int_\limits0^1 \sqrt{\frac{1-x}{1+x}} \mathrm{~d} x$$ is

$$\int_\limits0^2[x] \mathrm{d} x+\int_\limits0^2|x-1| \mathrm{d} x=$$

(where $$[x]$$ denotes the greatest integer function.)