Find the area under the graph of $f$ over the interval $[-1,3]$. \[ f(x)=\left\{\begin{array}{ll} x^{2}+1 & x \leq 1 \\ 2 x & x>1 \end{array}\right. \] 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 \(x \leq 1\) and one for \(x > 1\).

Step 2 ：For \(x \leq 1\), 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 \(\boxed{\frac{32}{3}}\).