Can I integrate an unknown?

Calculus Level 5

$\large \int_0^x \frac{f(t)}{\sqrt{x-t}} \, dt = 1+2x-x^2$

Find the value of $f(1)$ . Give your answer to 3 decimal places.

The answer is 0.7427.

1 solution

Aman Rajput
Jun 14, 2015

This is popularly known as ABEL's INTEGRAL EQUATION.

It can be written as :

$\displaystyle f(x)x^{\frac{-1}{2}} = 1 + 2x - x^2$

Taking Laplace transforms on both the sides and using convolution theorem , we get

$\displaystyle \bar{f}\mathcal{L}(x^{\frac{-1}{2}}) = \mathcal{L}(1 + 2x - x^2)$

$\displaystyle \bar{f} \frac{\Gamma(1/2)}{s^{1/2}} = \frac{1}{s} + \frac{2}{s^2} - \frac{2}{s^3}$

$\displaystyle \bar{f} = \frac{1}{\Gamma(1/2)}(\frac{1}{s^{1/2}} + \frac{2}{s^{3/2}} - \frac{2}{s^{5/2}})$

On inversion , and using $\displaystyle \mathcal{L}^{-1}\frac{1}{s^{n+1}} = \frac{x^n}{\Gamma(n+1)}$

we get finally,

$\displaystyle f(x)=\frac{1}{3\pi\sqrt{x}}(3 + 12x - 8x^2)$

Thus , $\displaystyle f(1) = 0.7428$

Thanks..!!! for reading the solution.

Very good solution

Vishesh Bansal - 5 years, 12 months ago

Interesting problem. I found that $\displaystyle f(x)=\frac{1}{3\pi\sqrt{x}}(3 + 12x - 8x^2),$ i.e. the term $x^3$ should be $x^2$ .

Jon Haussmann - 5 years, 9 months ago

no .. it is alright.

Aman Rajput - 5 years ago

Taking $n = 3/2$ in $\mathcal{L}^{-1}\dfrac{1}{s^{n+1}} = \dfrac{x^n}{\Gamma(n+1)}$ , we find that $\dfrac{2}{s^{5/2}}$ produces a term involving $x^{3/2} = \dfrac{x^2}{\sqrt{x}}$ , so $f(x) = \frac{1}{3 \pi \sqrt{x}} (3 + 12x - 8x^2).$

One can also check manually: If $f(x) = \frac{1}{3 \pi \sqrt{x}} (3 + 12x - 8x^3),$ then $\int_0^x \frac{f(t)}{\sqrt{x - t}} \ dt = 1 + 2x - \frac{5}{6} x^3.$ The function $f(x) = \frac{1}{3 \pi \sqrt{x}} (3 + 12x - 8x^\color{#D61F06}{2})$ gives the correct result of $1 + 2x - x^2$ .

Jon Haussmann - 5 years ago

@Jon Haussmann – Okay right. :)

Aman Rajput - 5 years ago

Can I integrate an unknown?

The answer is 0.7427.

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