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Question Number 31141 by Cheyboy last updated on 03/Mar/18

using the limit defination  find the area of  f(x)= cos(x)  [0,π/2]

$$\boldsymbol{{using}}\:\boldsymbol{{the}}\:\boldsymbol{{limit}}\:\boldsymbol{{defination}} \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{area}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\:\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\:\:\left[\mathrm{0},\pi/\mathrm{2}\right] \\ $$

Answered by Joel578 last updated on 03/Mar/18

A = ∫_0 ^(π/2)  cos x dx = lim_(n→∞)  Σ_(i=1) ^n  f(x_i )Δx_i        = lim_(n→∞)  Σ_(i=1) ^n  cos (((iπ)/(2n))) (π/(2n))       = lim_(n→∞)  (π/(2n)) Σ_(i=1) ^n  cos (((iπ)/(2n)))       = (π/2) lim_(n→∞)  ((Σ_(i=1) ^n  cos (((iπ)/(2n))))/n)       = (π/2) lim_(n→∞)  ((cos ((π/(2n))) + cos ((π/n)) + cos (((3π)/(2n))) + ... + cos (((nπ)/(2n))))/n)

$${A}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}\:{x}\:{dx}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:{f}\left({x}_{{i}} \right)\Delta{x}_{{i}} \\ $$$$\:\:\:\:\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{cos}\:\left(\frac{{i}\pi}{\mathrm{2}{n}}\right)\:\frac{\pi}{\mathrm{2}{n}} \\ $$$$\:\:\:\:\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\pi}{\mathrm{2}{n}}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{cos}\:\left(\frac{{i}\pi}{\mathrm{2}{n}}\right) \\ $$$$\:\:\:\:\:=\:\frac{\pi}{\mathrm{2}}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{cos}\:\left(\frac{{i}\pi}{\mathrm{2}{n}}\right)}{{n}} \\ $$$$\:\:\:\:\:=\:\frac{\pi}{\mathrm{2}}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}{n}}\right)\:+\:\mathrm{cos}\:\left(\frac{\pi}{{n}}\right)\:+\:\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{2}{n}}\right)\:+\:...\:+\:\mathrm{cos}\:\left(\frac{{n}\pi}{\mathrm{2}{n}}\right)}{{n}} \\ $$

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