Answer
a. $\frac{17}{6}$
b. $F(x) = \frac{x^{2}}{2}+C1, 0\leq x \leq 1$
$F(x) = \frac{x^{3}}{3}+C2, 1\lt x \leq 2$
Work Step by Step
a. $$\int_{0}^{2} f(x) dx = \int_{0}^{1} f(x) dx + \int_{1}^{2}f(x)dx = \int_{0}^{1} x dx + \int_{1}^{2}x^2dx = \frac{x^{2}}{2}]_{0}^{1}+ \frac{x^{3}}{3}]_{1}^{2} = (\frac{1}{2}-0)+(\frac{8}{3} - \frac{1}{3}) = \frac{1}{2}+\frac{7}{3} = \frac{17}{6} $$
b. anti-derivative is the indefinite integral of the original function, but since there is a piecewise function, we will have two F(x) depending on the range of x. We've calculated part of the anti-derivative above, and I will not go into power rules for this example. Just one thing to add, since function is continuous at x=1, we have to relate C1 with C2 by setting the piecewise equal to each other, which will come useful in the next part:
$$\frac{1^{2}}{2}+C1= \frac{1^{3}}{3}+C2$$
$$C2-C1= \frac{1}{2}-\frac{1}{3} = \frac{1}{6}$$
c. FTC part one state that $\int_{a}^{b}f(x)dx = F(b)-F(a)$. In our example, $$\int_{0}^{2}f(x)dx = F(2)-F(0) = \frac{2^{3}}{3}+C2 -(\frac{0^{2}}{2}+C1) = \frac{8}{3}+(C2-C1) = \frac{8}{3}+\frac{1}{6} = \frac{17}{6}$$
QED
