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Integral Calculus> If f(x)= (pcosx+qsinx)(x 2 +ax+b) given ∫...

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If f(x)= (pcosx+qsinx)(x²+ax+b)
given ∫f(x)dx from lower limit -π/2 to upper limit π/2 vanishes for all p,q ϵ R, then
the value of π²+4b+a = ….......... ?

Jayant Kishore , 10 Years ago

Grade 12

1 Answers

Arju mukherjee

f(x)=(pcosx + qsinx)(x²+ $\alpha$ x+ $\beta$ ) [Given]

$\int_{\frac{-\pi }{2}}^{\frac{\pi }{2}} f(x) dx =\int_{0}^{\frac{\pi }{2}}[ f(x)+f(-x)] dx$

$\Rightarrow \int_{0}^{\frac{\pi }{2}} [(p\cos x + q\sin x)(x^{2}+\alpha x+\beta ) +(p\cos x - q\sin x )(x^{2}-\alpha x+\beta )] dx$

$\Rightarrow 2p\int_{0}^{\frac{\pi }{2}} (x^{2}+\beta )\cos x dx+2\alpha q\int_{0}^{\frac{\pi }{2}} x\sin x dx$

$\Rightarrow 2\left [p(x^{2}+\beta )\sin x + \left ( q\alpha - 2p \right )\left ( \sin x-x\cos x \right ) \right ]_{0}^{\frac{\pi }{2}}$

$\Rightarrow 2\left [ p(\frac{\pi ^{2}}{4}+\beta ) + q\alpha - 2p \right ]$

$\Rightarrow 2\left [ \frac{(\pi ^{2}+4\beta )}{4} + q\alpha - 2p \right ]$

$\Rightarrow 2\left [\left (\frac{(\pi ^{2}+4\beta )}{4}-2 \right )p + q\alpha \right ]$

As it is vanishes for all value of p and q, co-efficient of p and q must be equal to 0.

Hence,

$\therefore 2\left (\frac{(\pi ^{2}+4\beta )}{4}-2 \right )=0 \Rightarrow (\pi ^{2}+4\beta )=8$

$\therefore 2\alpha =0\Rightarrow \alpha =0$

Hence,
$(\pi ^{2}+4\beta +\alpha )=8$

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