1sorry but thats not the f(x)
check ur soln once more
btw i hav a diff method
\boxed{1} Find the value of a for which \int_{0}^{\pi}\{ax(\pi^{2}-x^{2})-\sin x\}^{2}dx is minimized. (easy one) for prac.
\boxed{2} Let f(x) be a continuous function satisfying 
Find the value of k for which 
is maximized.
\\\boxed{3} \lim_{n\to\infty}\int_{0}^{n\pi}e^{-x}|\sin x| dx\\ \\\boxed{4} \lim_{n\to\infty}\int_{0}^{\frac{\pi}{2}}\frac{[n\sin x]}{n}\ dx \textup{ where [.] is gif}
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29The last post got deleted when i was trying to edit it..
btw posting that same stuff again...
f(x) = 1 + ka
where a = \int_{-\pi/2}^{\pi/2}{f(t) sin(x-t)}dt
Now {f(t) sin(x-t)}dt = sin(x-t) + ka sin(x-t)
Integrating both sides we get
\int_{-\pi/2}^{\pi/2}{{f(t) sin(x-t)}dt} = \int_{-\pi/2}^{\pi/2}sin(x-t) + ka \int_{-\pi/2}^{\pi/2}sin(x-t)
a = (1+ ka) \int_{-\pi/2}^{\pi/2}sin(x-t)dt = (1+ ka) \int_{-\pi/2}^{\pi/2}sinxcostdt - cosxsintdt
a = (1+ ka) (sinx\int_{-\pi/2}^{\pi/2}costdt - cosx\int_{-\pi/2}^{\pi/2}sintdt )
a = (1+ ka) 2sinx \Rightarrow cosecx = \frac{2}{a} + 2k \Rightarrow cosecx-2k = \frac{2}{a}
a = \frac{2}{cosecx -2k} \Rightarrow f(x) = 1 + \frac{2k}{cosecx- 2k}
now how to solve it furthe..
662) We have
f(x)=1+k\sin x\int_{-\pi/2}^{\pi/2}f(t)\cos t\ \mathrm dt -k\cos x\int_{-\pi/2}^{\pi/2}f(t)\sin t\ \mathrm dt
That is
f(x)=1+ka\sin x -kb\cos x
where
a=\int_{-\pi/2}^{\pi/2}f(t)\cos t\ \mathrm dt ----(1)
and
b=\int_{-\pi/2}^{\pi/2}f(t)\sin t\ \mathrm dt ---(2)
We then have
\int_0^\pi f(x)\ \mathrm dx=\pi + 2ka ----- (3)
On the other hand from (1), we get
a=\int_{-\pi/2}^{\pi/2}(\cos t+ka\sin t \cos t-kb\cos^2t)\ \mathrm dt=2-\dfrac{bk\pi}{2}
That is
a+\dfrac{bk\pi}{2}=2 ----- (4)
And from (2), we get
b=\int_{-\pi/2}^{\pi/2}(\sin t+ka\sin^2t-kb\cos t\sin t)\ \mathrm dt=\dfrac{ak\pi}{2} ---- (5)
Solving (4) and (5) for a and b, we get
a=\dfrac{2}{1+\left(\frac{k\pi}{2}\right)^2}
and b=\dfrac{k\pi}{1+\left(\frac{k\pi}{2}\right)^2}
Hence
\int_0^\pi f(x)\ \mathrm dx = \pi + \dfrac{8}{\pi}\cdot\dfrac{\frac{k\pi}{2}}{1+\left(\frac{k\pi}{2}\right)^2}
which maximizes when
\dfrac{k\pi}{2}=1\quad\Rightarrow\ \boxed{k=\dfrac{2}{\pi}}
1kaymant sir as always, absolutely right [1]
anyone finishing the first one except experts ?
1ans for 3rd is \frac{1}2\cdot\frac{e^{\pi}+1}{e^{\pi}-1}
ans for 4th is 1
btw 3rd is ma doubt as well... :P
66Q3) We have
I=\int_0^{n\pi}e^{-x}|\sin x|\ \mathrm dx =\sum_{k=0}^{n-1}(-1)^k \int_{k\pi}^{(k+1)\pi}e^{-x}\sin x\ \mathrm dx
The integral
\int_{k\pi}^{(k+1)\pi}e^{-x}\sin x\ \mathrm dx =\left(\dfrac{1+e^{-\pi}}{2}\right)(-e^{-\pi})^k
Hence,
I=\left(\dfrac{1+e^{-\pi}}{2}\right)\sum_{k=0}^{n-1}(e^{-\pi})^k=\dfrac{1}{2}\cdot \dfrac{1+e^{-\pi}}{1-e^{-\pi}}\cdot\left(1-e^{-n\pi}\right)
Now taking the limit as n→∞, we get finally that
I=\left(\dfrac{1+e^{-\pi}}{2}\right)\sum_{k=0}^{n-1}(e^{-\pi})^k=\dfrac{1}{2}\cdot \dfrac{1+e^{-\pi}}{1-e^{-\pi}}