Compute $\int _{\frac{4}{5}}^2\:f^{-1}\left(x\right)dx$ where $f\left(x\right)=\frac{-x^3+2x^2-5x+8}{x^2+4},\:x\in \mathbb{R}$ is bijection.

Question

We have to compute $\int _{\frac{4}{5}}^2\:f^{-1}\left(x\right)dx$ where $f\left(x\right)=\frac{-x^3+2x^2-5x+8}{x^2+4},\:x\in \mathbb{R}$ is an bijective function.

What we can extract from bijective information?

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EDIT : Based on Answers, I tried to Plot this

Answer 1

Hint: If the function bijective, you can draw the function as I drew below

Step 1: Find $a$ and $b$

$f(a)=\frac{4}{5} $

$f(b)=2 $

And compute $B=\int _{a}^b\:f\left(x\right)dx$

Step 2:

Compute rectangle area of C and compute $B-C$

Step 3:

$A=\int _{\frac{4}{5}}^2\:f^{-1}\left(x\right)dx$

$A+B-C$ is also a rectangle .

Answer 2

You may notice that $f(x)$ is a decreasing function on the interval $[0,1]$ and $f(0)=2,\,f(1)=\frac{4}{5}$, from which it follows that (just look at the picture below to figure it out): $$ \int_{\frac{4}{5}}^{2}f^{-1}(x)\,dx = -\frac{4}{5}+\int_{0}^{1}f(x)\,dx=\color{red}{\frac{7}{10}+\log 2-\frac{\log 5}{2}}\tag{1}$$ where the last result follows from polynomial division.

Answer 3

draw a graph of $y = f(x).$ you will see that it has $y$-intercept at $(0,1)$ and has a point $(1,4/5).$ the area you are after is the triangular region bounded by $x = 0$ on the left, by $y = \frac 45$ at the bottom and $y = f(x)$ like a hypotenuse. the area $$\int_{4/5}^2 x\,dy$$ is like cutting the triangle in slices/strips parallel to the $x-$ axis. now, we can find the same area by slicing parallel to the $y$-axis. this way we get the area to be $$ \int_0^1\left(y - \frac 45\right)\, dx$$

this is what Mathlover was trying to show you with a picture. hope this is of some use to you.

Answer 4

Let $f(u)=x$, then integration by parts gives $$ \begin{align} \int_{4/5}^2f^{-1}(x)\,\mathrm{d}x &=\int_1^0u\,\mathrm{d}f(u)\\ &=-\int_0^1u\,\mathrm{d}f(u)\\ &=-\left[uf(u)\vphantom{\int}\right]_0^1+\int_0^1f(u)\,\mathrm{d}u\\ &=-\frac45+\int_0^1\frac{-u^3+2u^2-5u+8}{u^2+4}\,\mathrm{d}u\\ &=-\frac45+\int_0^1\frac{2u^2+8}{u^2+4}\,\mathrm{d}u-\int_0^1\frac{u^2+5}{u^2+4}u\,\mathrm{d}u\\ &=\frac65-\frac12\int_4^5\frac{v+1}v\,\mathrm{d}v\\ &=\frac7{10}-\frac12\log\left(\frac54\right) \end{align} $$ where $v=u^2+4$.