Find the area under curve using the given data

Question

Consider the curve $y=f(x)$ which satisfies the DE $(1+x^2)\frac{dy}{dx} +2xy=4x^2$ and passes through the origin. Find area enclosed by $f^{-1}x$, x axis, and $x=2/3$

After some calculation which I don’t think is necessary to show here, I got $$y=\frac{4x^3}{3(1+x^2)}$$

Now the inverse for this function can’t be found directly, and I don’t know how else to do it. Can I get a hint?

Answer 1

we have $$I=\int_{0}^{2/3} f^{-1}(x)dx$$ take $x=f(t),dx=f'(t)dt$ then $$I=\int_{0}^1 tf'(t)dt=1\cdot f(1)-0\cdot f(0)-\int_0^1 f(t) \, dt$$ can you finsish?

Answer 2

Notice that $f(1) = \frac{2}{3}$ and $f(0) = 0$

Thus the integrals of the function and its inverse will add up to the rectangle of area $\frac{2}{3}\cdot 1$ as we can see on this graph

$$\int_0^{\frac{2}{3}}f^{-1}(x)dx+\int_0^1f(x)dx = \frac{2}{3}$$

The antiderivative of $f$ is given by

$$\frac{2}{3}\int 2x\left(1-\frac{1}{1+x^2}\right)dx = \frac{2}{3}x^2-\frac{2}{3}\ln(1+x^2)+C$$

which means the final answer is

$$\frac{2}{3}\ln 2$$

Answer 3

Since you received very good and simple answers, let me be stupid and work with the inverse function.

$$y=\frac{4x^3}{3(1+x^2)}\implies 4 x^3-3 y\,x^2-3 y=0$$ Using the hyperbolic method for solving cubic equations when only one real root, $$x=\frac{1}{4} y \left(1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(1+\frac{24}{y^2}\right)\right)\right)$$ which does not show any antiderivative !