Evaluate this definite integral: $$\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$$
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1One evaluates or calculates integrals, rather than solving them. One solves equations, problems, &c. – Pedro May 25 '13 at 01:04
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Right! I'm sorry for the mistake, will keep that in mind. Thanks! – Soham Chowdhury May 25 '13 at 13:50
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This integral (even the indefinite one) can be easily solved by observing:
$$\frac{\mathrm d}{\mathrm dx}\pi\sqrt x = \frac{\pi}{2\sqrt x}$$
which implies that:
$$\frac{\mathrm d}{\mathrm dx}\cos\pi\sqrt x = -\frac{\pi \sin\pi\sqrt x}{2\sqrt x}$$
Finally, we obtain:
$$\int\frac{\pi\sin\pi\sqrt x}{\sqrt x}\,\mathrm dx = -2\cos\pi\sqrt x$$
whence the definite integral with bounds $0, n^2$ evaluates to $2(1-(-1)^n)$.
Lord_Farin
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$$\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$$ $$\implies\pi\int^{441}_0\frac{\sin \pi \sqrt x}{\sqrt x} dx$$ $$\implies\pi\int^{21}_0\frac{2u \sin \pi u}{u} du$$ $$\implies2\pi\int^{21}_0 \sin{\pi u}\;du$$ $$\implies2\pi\int^{21\pi}_0 \frac{\sin{v}}{\pi}\;dv$$ $$\implies2\int^{21\pi}_0 \sin{v}\;dv$$ $$\implies -2 \left[\cos \pi \sqrt x\right]^{21}_{0}$$ $$\implies-2(\cos 21\pi - \cos 0)$$ $$\implies-2(-1 - 1)$$ $$\implies-2\times (-2)$$ $$\implies 4$$
An interesting thing is that for upper bounds that are squares of even numbers this evaluates to zero, while for squares of odd numbers (such as $21^2 = 441$) this evaluates to 4.
Jon Claus
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Soham Chowdhury
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I guess I would have made a single substitution of $ u = \pi \sqrt{x} $ so as to limit the number of times substitution has to be used, but yes that looks good regardless. As for the mention of the evenness and oddness of the resulting number just replace the upper bound with the number $n^2$ and you get as the resulting answer (when evaluated from zero to n as per the substitution) $$ -2 (\cos(n \pi) - 1 ) = -2 ( (-1)^n - 1)$$ which is non zero when n is odd and zero when n is even. – Triatticus May 23 '13 at 04:11
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