Need help in solving defined integral problem.

Question

The graph of the continuous function $y = f (x)$ is symmetric with respect to the origin, for all real numbers $x$

If $$ f(x)=\frac{\pi}{2}\int_1^{x+1}f(t)dt$$ and $f(1)=1$

Find $$\pi^2\int_0^1xf(x+1)dx$$ I have tried:$$f'(x)=\frac{\pi}{2}f(x+1)$$ and as the function is odd $f(-1)=-1$ $$\frac{\pi}{2}\int_0^1f(x+1)dx=1$$
But I can`t go any further can someone help me with it?

score 1 · Accepted Answer · answered Jul 19 '17 at 09:10

1

From given condition, $$f(0) = 0, \ \ f'(x) = \frac{\pi}{2}f(x+1), \ \ f'(x-1) = \frac{\pi}{2}f(x) $$

Then,

$$\begin{align}\pi^2\int_{0}^{1}xf(x+1)dx &= 2\pi\int_{0}^{1}xf'(x)dx \\\\ &= 2\pi(xf(x)|_{0}^{1} - \int_{0}^{1}f(x)dx) \\\\ &= 2\pi - 4\int_{0}^{1}f'(x-1)dx \\\\ &= 2\pi - 4f(x-1)|_{0}^{1} \\\\ &= 2\pi + 4f(-1) \\\\ &= 2\pi-4 \end{align}$$

answered Jul 19 '17 at 09:10

Dhruv Kohli

5,216

This is being very helpful but can you clarify how you changed f(x) in the defined integral from o to 1 in the third line to 4 times defined integral from 0 to 1 of f '(x-1) – R.Temur Jul 19 '17 at 09:20
$f'(x-1) = \frac{\pi}{2}f(x)$ – Dhruv Kohli Jul 19 '17 at 09:21
@R.Temur Did you understand? – Dhruv Kohli Jul 19 '17 at 09:27
from where f′(x−1)=π/2f(x) was taken or found – R.Temur Jul 19 '17 at 09:30
how to find it , please help – R.Temur Jul 19 '17 at 09:31
Replace $x$ by $x-1$ in the given equation and then take derivative wrt $x$. Got it? – Dhruv Kohli Jul 19 '17 at 09:33
Let us continue this discussion in chat. – R.Temur Jul 19 '17 at 10:07

Need help in solving defined integral problem.

1 Answers1