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A function $f(x)$, where $x$ is a real number , is defined implicitly by the following formula: $$f(x)=x-\int^{\frac{\pi}{2}}_0f(x)\sin(x)dx$$ Find the explicit function for $f(x)$ in its simplest form.

This question appeared in the recent New Zealand Qualifications Authority 2016 Scholarship Calculus examination.


What I have done

Let $f(x)=y$

$$f(x)=x-\int^{\frac{\pi}{2}}_0f(x)\sin(x)dx \Rightarrow y=x-\int^{\frac{\pi}{2}}_0y\sin(x)dx$$

$$ y=x-\int^{\frac{\pi}{2}}_0y\sin(x)dx \Leftrightarrow \int^{\frac{\pi}{2}}_0y\sin(x)dx=x-y$$

Consider the integral

$$ \int^{\frac{\pi}{2}}_0y\sin(x)dx$$

Let $u=y\Rightarrow du=dy$ and $dv= \sin(x) \Rightarrow v=-\cos(x)$

$$ \int^{\frac{\pi}{2}}_0y\sin(x)dx = \left[y\sin(x) \right]^{\frac{\pi}{2}}_0+\int^{\frac{\pi}{2}}_0\cos(x) dydx$$

How can I continue?

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  • $\begingroup$ The use of $x$ in the integral and outside of the integral is somewhat hard to read. $\endgroup$ – Simply Beautiful Art Nov 25 '16 at 1:58
  • $\begingroup$ @SimpleArt This is how the question was worded , sorry. $\endgroup$ – bigfocalchord Nov 25 '16 at 2:41
  • $\begingroup$ @Joseph , haha how did you know it was from the NCEA exam? $\endgroup$ – bigfocalchord Nov 25 '16 at 2:44
  • $\begingroup$ @dydxx I just sat it! :) I came looking for some pointers, and this question was the first one I saw. $\endgroup$ – Jollywatt Nov 25 '16 at 2:47
  • $\begingroup$ @Joseph I found it extremely hard compared to last years paper, conics killed me haha. $\endgroup$ – bigfocalchord Nov 25 '16 at 2:50
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Hint. The integral in your equation is a definite integral and evaluates to a constant. Therefore $$f(x)=x-c$$ where $c$ is a constant satisfying $$c=\int_0^{\pi/2} (x-c)\sin x\,dx\ .$$ Evaluate the integral, then solve the equation to find $c$.

Answer:

$c=\frac12$.

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$$f(x)=x-\int^{\frac{\pi}{2}}_0f(x')\sin(x')dx'$$ multiply by $\sin(x)$ and integrate from $0$ to $\frac{\pi}{2}$: $$f(x)\sin(x)=x\sin(x)-\left(\int^{\frac{\pi}{2}}_0f(x)\sin(x)dx\right)\sin(x)$$ $$\int^{\frac{\pi}{2}}_0f(x)\sin(x)dx=\int^{\frac{\pi}{2}}_0x\sin(x)dx-\left(\int^{\frac{\pi}{2}}_0f(x)\sin(x)dx\right)\int^{\frac{\pi}{2}}_0\sin(x)dx$$ $$x-f(x)=1-(x-f(x))(1)$$ So $$2(x-f(x))=1$$ which means $$f(x)=x-\frac{1}{2}$$

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  • $\begingroup$ Be careful. You should not use $x$ as both the variable of integration and a variable outside the integral. They are different things and should be named differently. $\endgroup$ – marty cohen Nov 25 '16 at 2:19
  • $\begingroup$ Agreed, but the first equation was from the question... $\endgroup$ – msm Nov 25 '16 at 2:25

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