Inversion Game Real Strong

$\begin{aligned} I & = \int_{2+\sin 1}^{4+\sin 2} f^{-1}(y) \ dy \\ & = \int_{2+\sin 1}^{4+\sin 2} x dy & \small \color{#3D99F6} \text{Note that }\frac {dy}{dx} = \cos x + 2 \\ & = \int_1^2 x(\cos x + 2) \ dx \\ & = {\color{#3D99F6} \int_1^2 x\cos x \ dx} + \int_1^2 2x \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = {\color{#3D99F6} x\sin x - \int \sin x dx} + x^2 \ \bigg|_1^2 \\ & = {\color{#3D99F6} x\sin x + \cos x} + x^2 \ \bigg|_1^2 \\ & = 2 \sin 2 - \sin 1 + \cos 2 - \cos 1 + 4 - 1 \\ & \approx \boxed{3.021} \end{aligned}$

Relevant wiki: Integration Tricks

$\begin{aligned} \displaystyle I & = \int_{2 + \sin 1}^{4 + \sin 2} f^{-1}(x) \ dx \\ & = \int_{f(1)}^{f(2)} f^{-1}(x) \ dx & \small \color{#3D99F6} \text{Use the identity }\int_a^b f(x) \ dx + \int_{f(a)}^{f(b)} f^{-1}(x) \ dx = bf(b) - af(a) \\ && \small \color{#3D99F6} \text{for increasing one-to-one functions} \\ & = 2f(2) - 1f(1) - \int_1^2 f(x) \ dx \\ & = 2\big(4 + \sin 2\big) - \big(2 + \sin 1\big) - \int_1^2 \sin x + 2x \ dx \\ & = 6 + 2\sin 2 - \sin 1 - \Bigg[x^2 - \cos x \bigg|_1^2\Bigg] \\ & = 3 + 2\sin 2 - \sin 1 + \cos 2 - \cos 1 \\ & \approx \boxed{3.021} \end{aligned}$

From inspection, we have $f(1)=2+\sin 1$ and $f(2)=4+\sin 2$ .

NOTE: We don't need a graphing calculator for solving this question. We can solve this question by just graphing it on a paper. I've used Desmos Graphing Calculator so that a better view of the graph can be provided.

Let $y=f(x)$ , then $x=f^{-1}\left(y\right)$ . We represent this on the x-y plane like this :

Thereby, we can conclude that $\large \displaystyle \int _{2+\sin \left(1\right)}^{4+\sin \left(2\right)}f^{-1}\left(y\right)dy$ (which is the same as $\large \displaystyle \int _{2+\sin \left(1\right)}^{4+\sin \left(2\right)}f^{-1}\left(x\right)dx$ ) is the area of the region $CDFE$ . Furthermore, $OBFE$ and $OADC$ are rectangles and the area of the region $ADFB$ is simply the integral $\large \displaystyle \int _1^2\sin \left(x\right)+2x\ dx$ .

We also have $ar(OADC) + ar(ADFB) + ar(CDFE) = ar(OBFE)$

$\large \displaystyle \implies 1(2+\sin 1) + ( \int _1^2\sin \left(x\right)+2x\ dx) + (\int _{2+\sin \left(1\right)}^{4+\sin \left(2\right)}f^{-1}\left(x\right)dx) = 2(4 + \sin 2)$

$\large \displaystyle \implies 2+\sin 1 + \cos 1 - \cos 2 + 3 + (\int _{2+\sin \left(1\right)}^{4+\sin \left(2\right)}f^{-1}\left(x\right)dx) = 8 + 2\sin 2$

$\large \displaystyle \implies \int _{2+\sin \left(1\right)}^{4+\sin \left(2\right)}f^{-1}\left(x\right)dx = 3\ +\ 2\sin 2-\sin 1-\cos 1+\cos 2 = \boxed{3.021}$