Let $f(t)$ be defined by $ f(t) = \left\{ \begin{array}{c} t, &0 \le t < 1 \\ b - t^2, &1 \le t \le 2 \end{array} \right. $ Let $f(t)$ be defined by
$$
f(t) = 
\left\{
\begin{array}{c}
t, &0 \le t < 1 \\ 
b - t^2, &1 \le t \le 2
\end{array}
\right.
$$
and let $F(x)$ be defined by $F(x) = \int_0^xf(t)dt,\,\,\,\,\, 0 \le x \le 2.$
a) Find F(x)
b) For what value of $b$ in the definition of $f$ is $F(x)$ differentiable for all $x \in [0,2]$?
What I have so far is..
if $0 \le t < 1$, then $\int_0^xt dt = \frac{x^2}{2}$
if $1 \le t \le 2$, then $\int_0^1t dt + \int_1^xb - t^2dt = \frac{-x^3}{3} +xb - b + \frac{5}{6}$
therefore
$
F(x) = 
\left\{
\begin{array}{c}
\frac{x^2}{2}, &0 \le t < 1 \\ 
\frac{-x^3}{3} +xb - b + \frac{5}{6}, &1 \le t \le 2
\end{array}
\right.
$
I'm not sure if I did A correctly and I need help with B.

This seems correct to me. Now you must check for which $b$ the right and left derivatives at $x = 1$ are equal. The left derivative is $1$ and the right is $-1+b$ so $1 = -1+b$ iff $b =2$.
