Hello, Susie38!
Here are parts (b) and (c).
$\displaystyle f(x) \:= \:\begin{Bmatrix}\frac{2}{\pi}x & 0 \leq x \leq \frac{\pi}{2} \\ 2\sin x & \frac{\pi}{2} < x \leq \frac{3\pi}{2} \\ 1 & \frac{3\pi}{2} < x \leq 2\pi\end{Bmatrix}$
a. Find a continuous function, $\displaystyle F(x)$, satisfying $\displaystyle F(x) = \int^x_0 f(t)\,dt$
b. Give the values of $\displaystyle x$ on $\displaystyle [0,2\pi]$ for which $\displaystyle F(x)$ is not differentiable.
c. Compute: $\displaystyle \int^{2\pi}_0 f(x)\,dx$
On $\displaystyle \left[0,\,\frac{\pi}{2}\right]\]$, the function is a straight line.
On $\displaystyle \left(\frac{\pi}{2},\,\frac{3\pi}{2}\right]$, the function is a sine curve.
On $\displaystyle \left(\frac{3\pi}{2},\,2\pi\right]$, the function is a horizontal line. Code:

1 + ** o * * * *
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 *    +    *    +    + 
 π/2 2π* 3π/2 2π
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1 + **

(b) $\displaystyle F(x)$ is not differentiable at $\displaystyle x = \frac{\pi}{2}$ and $\displaystyle x = \frac{3\pi}{2}$
It has two different slopes at $\displaystyle x = \frac{\pi}{2}$
It is discontinuous at $\displaystyle x = \frac{3\pi}{2}$
(c) No calculus needed!
The triangle has base $\displaystyle \frac{\pi}{2}$ and height $\displaystyle 1.$
. . . Its area is: $\displaystyle A_1 \:= \:\frac{1}{2}\left(\frac{\pi}{2}\right)(1)\:=\; \frac{\pi}{4}$
The sine curve has a region above and a region below the xaxis.
. . . The two regions have equal area and will cancel out.
Its net area is: $\displaystyle A_2\:=\:0$
The rectangle has base $\displaystyle \frac{\pi}{2}$ and height $\displaystyle 1.$
. . . Its area is: $\displaystyle A_3\:=\:\left(\frac{\pi}{2}\right)(1)\:=\: \frac{\pi}{2}$
The value of the integral is the sum of these areas:
. . . $\displaystyle A_1 + A_2 + A_3 \;= \;\frac{\pi}{4} + 0 + \frac{\pi}{2}\;=\;\frac{3\pi}{4}$