Hello, Tulki!
Here's part (a).
Did you make a sketch?
$\displaystyle \text{Let }\:f(x) \;=\;\left\{\begin{array}{cccc}0 && x < 0 \\ \\[4mm]
x && 0 \leq x \leq 1 \\ \\[4mm]
2  x && 1 < x \leq 2 \\ \\[4mm]
0 && x > 2 \end{array}\right\}$
and: .$\displaystyle g(x) \:=\: \int_0^x f(t)\,dt$
a) Find an expression for $\displaystyle g(x)$ similar to the one for $\displaystyle f(x).$
$\displaystyle g(x)$ gives the area under the graph from $\displaystyle 0\text{ to }x$
The graph looks like this: Code:


1+ *
 *:::*
 *:::::::*
 *::::::::: *
 * * *    *  +  * * *  
0 1 x 2
Then: .$\displaystyle \displaystyle g(x) \;=\;\left\{ \begin{array}{cccc}
0 && x < 0 \\ \\[3mm]
\int^x_0t\,dt && 0 \leq x \leq 1 \\ \\[2mm]
\frac{1}{2} + \int^x_1 (2t)\,dt && 1 < x \leq 2 \\ \\[3mm]
1 && x > 2
\end{array} \right\} $