Hi William,

For #17 you are correct, the answer (rounded to the nearest hundredth) should be $14.29$. This has been fixed in the quiz so now it should be marked as correct. Thanks for pointing this out!

For #20 we need to split the integral in two portions, since the shaded region is bounded by $f(x)$ and the $x$-axis on the first portion and by $g(x)$ and the $x$-axis on the second portion. The graphs of $f(x) = x^2 + 1$ and $g(x) = -x^2 + 9$ intersect when $x = 2$, so we want to find $$\int_{1}^{2} (x^2 + 1)\, dx + \int_{2}^{3} (-x^2 + 9) \,dx.$$ The first integral is equal to $$\int_{1}^{2} (x^2 + 1) \,dx = \left(\dfrac{x^3}{3} + x\right)\mid_{1}^{2} = \dfrac{10}{3},$$ and the second integral is equal to $$\int_{2}^{3} (-x^2 + 9) \,dx = \left(-\dfrac{x^3}{3} + 9x \right)\mid_{2}^{3} = \dfrac{8}{3},$$ so the area bounded by those graphs and the $x$-axis  is $\dfrac{10}{3} + \dfrac{8}{3} = 6$.

Let us know if you have any other questions!