Let f(x) = x³ + x. Since f'(x) = 3x² + 1 > 0, the function f increases on the whole real line, and…

(a) $f(2) = 8 + 2 = 10$, so $f^{-1}(10) = 2$.

(b) The region beside the curve, with $0 \le y \le 10$, has area $\displaystyle\int_0^{10} f^{-1}(y)\,dy$ (horizontal strips of length $x = f^{-1}(y)$). With the region under the curve it fills the $2 \times 10$ rectangle. The region under the curve has area $\displaystyle\int_0^2 (x^3 + x)\,dx = 4 + 2 = 6$. So the integral is $20 - 6 = 14$.

(c) $f(1) = 2$, so $y = 2$ corresponds to $x = 1$. The region $\{2 \le y \le 10,\ 0 \le x \le f^{-1}(y)\}$ is the rectangle $[0, 2] \times [0, 10]$, less the rectangle $[0, 1] \times [0, 2]$, less the area under the curve for $1 \le x \le 2$: $20 - 2 - \left(\dfrac{15}{4} + \dfrac32\right) = \frac{51}{4} = 12.75$.

(d) Proof. First, for any $c > 0$: $\displaystyle\int_0^{c} \varphi(x)\,dx + \int_0^{\varphi(c)} \varphi^{-1}(y)\,dy = c\,\varphi(c)$. To see it, put $y = \varphi(x)$ in the second integral: it becomes $\displaystyle\int_0^c x\,\varphi'(x)\,dx = c\,\varphi(c) - \int_0^c \varphi(x)\,dx$, by parts. Now fix $b > 0$ and let $h(a) = \displaystyle\int_0^a \varphi(x)\,dx + \int_0^b \varphi^{-1}(y)\,dy - ab$ for $a > 0$. Then $h'(a) = \varphi(a) - b$. Since $\varphi$ increases from $0$ without bound, there is exactly one $a_0 > 0$ with $\varphi(a_0) = b$; $h' < 0$ before it and $h' > 0$ after it. So $h(a) > h(a_0)$ for every $a \ne a_0$, and $h(a_0) = \displaystyle\int_0^{a_0}\varphi + \int_0^{\varphi(a_0)} \varphi^{-1} - a_0\varphi(a_0) = 0$ by the first step. So $h(a) \ge 0$, with equality only when $a = a_0$, that is, when $b = \varphi(a)$.

(e) For $\varphi(x) = x^2$, $\varphi^{-1}(y) = \sqrt y$, and (d) says $ab \le \dfrac{a^3}{3} + \dfrac23 b^{3/2}$. With $b = 9$: $9a \le \dfrac{a^3}{3} + 18$, so $\dfrac{a^3}{3} - 9a \ge -18$, with equality when $9 = a^2$, that is $a = 3$. The least value is -18.