Find the area under the graph of $f$ over the interval $[-1,3]$.
$$f(x)=\{\begin{array}{ll}{x}^2+1& x\le 1\\ 2x& x>1\end{array}$$
The area is (Simplify your answer.)

Step 1 ：The function is defined differently for different parts of the interval, so we will need to split the integral into two parts: one for $x\le 1$ and one for $x>1$.

Step 2 ：For $x\le 1$, the function is $f(x)=x^2+1$. We integrate this from -1 to 1 to find the area under the curve in this interval.

Step 3 ：The integral of $f(x)=x^2+1$ from -1 to 1 is $\frac{8}{3}$.

Step 4 ：For $x>1$, the function is $f(x)=2x$. We integrate this from 1 to 3 to find the area under the curve in this interval.

Step 5 ：The integral of $f(x)=2x$ from 1 to 3 is 8.

Step 6 ：We add these two areas together to find the total area under the curve from -1 to 3.

Step 7 ：The total area under the curve from -1 to 3 is $\frac{8}{3}+8=\frac{32}{3}$.

Step 8 ：Final Answer: The area under the graph of $f$ over the interval $[-1,3]$ is $\boxed{{\displaystyle \frac{32}{3}}}$.