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Question Number 11865 by ankhaaankhaa last updated on 03/Apr/17

∫_2 ^4 (√(16−x^2 ))dx/x^4 =

$$\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }{dx}/{x}^{\mathrm{4}} = \\ $$

Answered by ajfour last updated on 03/Apr/17

I= ∫_2 ^4 ((√(16−x^2 ))/x^4 )dx = ∫_2 ^4 ((√(((16)/x^2 )−1))/x^3 ) dx now let ((16)/x^2 )=t ⇒ ((−32dx)/x^3 ) =dt Further t=1 when x=4 and t= 4 when x=2 Then I= −(1/(32))∫_4 ^1 (√(t−1)) dt = (1/(32))∫_1 ^4 (√(t−1)) dt = (1/(32))(((t−1)^(3/2) )/((3/2))) ∣_1 ^4 = (1/(48)) (3(√3) ) = ((√3)/(16)) .

$${I}=\:\int_{\mathrm{2}} ^{\mathrm{4}} \frac{\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{4}} }{dx}\:=\:\int_{\mathrm{2}} ^{\mathrm{4}} \frac{\sqrt{\frac{\mathrm{16}}{{x}^{\mathrm{2}} }−\mathrm{1}}}{{x}^{\mathrm{3}} }\:{dx} \\ $$$${now}\:{let}\:\frac{\mathrm{16}}{{x}^{\mathrm{2}} }={t}\:\:\Rightarrow\:\:\frac{−\mathrm{32}{dx}}{{x}^{\mathrm{3}} }\:={dt} \\ $$$${Further}\:{t}=\mathrm{1}\:{when}\:{x}=\mathrm{4} \\ $$$$\:\:\:\:\:\:{and}\:\:{t}=\:\mathrm{4}\:\:{when}\:\:{x}=\mathrm{2}\: \\ $$$${Then}\:{I}=\:−\frac{\mathrm{1}}{\mathrm{32}}\int_{\mathrm{4}} ^{\mathrm{1}} \sqrt{{t}−\mathrm{1}}\:{dt} \\ $$$$\:\:\:\:\:=\:\:\frac{\mathrm{1}}{\mathrm{32}}\int_{\mathrm{1}} ^{\mathrm{4}} \:\sqrt{{t}−\mathrm{1}}\:{dt} \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{32}}\frac{\left({t}−\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }{\left(\mathrm{3}/\mathrm{2}\right)}\:\mid_{\mathrm{1}} ^{\mathrm{4}} \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{48}}\:\left(\mathrm{3}\sqrt{\mathrm{3}}\:\right)\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{16}}\:\:. \\ $$

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