Area enclosed by $f(x) = x^2$; $g(x) = \sqrt x$; and $h(x) = 6-x$

Question

I understand the basic concept of finding area between curves but the answer to this problem is supposedly 11/3 but I am stuck with a crazy fraction.

Please help.

Answer 1

This is one of those ambiguously framed questions which teachers should beware of.

Actually, $\dfrac{11}{3}$ is the area of the darker shaded region in the illustration below. But the entire shaded region is the region bounded by the three graphs, so $\dfrac{11}{3}$ is an incorrect answer.

The correct answer is found as follows:

\begin{equation} \int_{-3}^1 6-x-x^2dx+\int_1^46-x-\sqrt{x}\,dx=\frac{56}{3}+\frac{35}{6}=\frac{49}{2} \end{equation}

Furthermore, one could argue about the tear drop shaped region at the bottom since $y=6-x$ does not form part of its boundary. Another reason that the problem is somewhat ambiguous.

Graph by desmos.com