Q.Prove that $\int_0^{2\pi} |\cos x|\,dx = 4$

By symmetry $\int_0^{2\pi}|\sin x|\,dx=2\int_0^{\pi}\sin x\,dx=4$.

On $[0,\pi]$, $\sin x\ge 0$; on $[\pi,2\pi]$, $\sin x\le 0$ so $|\sin x|=-\sin x$. The two humps have equal area, so

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Even function: $\int_{-2}^2|x|\,dx=2\int_0^2 x\,dx=4$.

$|x|$ is even, so $\int_{-2}^{2}|x|\,dx=2\int_0^2|x|\,dx=2\int_0^2 x\,dx$ (since $x\ge0$ on $[0,2]$).

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$\dfrac{|x|}{x} = -1$ for $x<0$ and $+1$ for $x>0$, an odd function, so the integral over the symmetric interval $[-1,1]$ is $0$ — option (B).

On $[1,2)$ the greatest-integer function gives $[x]=1$, so the integral becomes a simple exponential integral.

For $x \in [1,2)$, the integer part $[x] = 1$ (the single point $x=2$ does not affect the value of a definite integral). So on this interval:

$e^{x+[x]} = e^{x+1}$

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On $[1,3]$, $x-3\le 0$, so $|x-3|=3-x$; integrate this directly.

For $x \in [1,3]$, $x \le 3$, so $x-3 \le 0$ and $|x-3| = 3-x$.

$\int_{1}^{3}|x-3|\,dx = \int_{1}^{3}(3-x)\,dx = \left[3x-\dfrac{x^2}{2}\right]_{1}^{3}$

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Split the integral at the integer break-point, since $[x]$ (greatest integer function) is a different constant on each sub-interval.

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Since the interval $[2,3]$ contains only positive numbers, $|x|=x$ there, reducing this to a standard power-rule integral.

For $x\in[2,3]$, $x>0$, so $|x|=x$. Thus

$\int_2^3 |x|\,dx = \int_2^3 x\,dx = \left[\frac{x^2}{2}\right]_2^3 = \frac{9}{2}-\frac{4}{2} = \frac{5}{2}$

Check: this is the area of a trapezium with parallel sides $2$ and $3$, width $1$: $\frac{(2+3)}{2}\times1=\frac52$ correct.

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The integral $\int_{1}^{3} |2x - 1|\, dx$ is evaluated by noting that $2x-1$ is positive on the entire interval $[1,3]$, so the absolute value drops directly. The result is $\boxed{6}$.

Part (a): $\displaystyle\int_{-2}^{2}|x|\,dx=4$. Part (b): $\displaystyle\int\frac{dx}{9+4x^2}=\frac16\tan^{-1}\!\left(\frac{2x}{3}\right)+C$.