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IntegrationQuestion and Answers: Page 26

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∫_0 ^(π/2) (((tan x))^(1/3) /(1+sin 2x)) dx =?

$$\:\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{tan}\:\mathrm{x}}}{\mathrm{1}+\mathrm{sin}\:\mathrm{2x}}\:\mathrm{dx}\:=? \\ $$

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∫_0 ^( 1) ∫_0 ^( 1) ∫_0 ^( 1) ((√(x + y + z))/( (√x) + (√y) + (√z) )) dxdydz

$$\: \\ $$$$\: \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\sqrt{{x}\:+\:{y}\:+\:{z}}}{\:\sqrt{{x}}\:+\:\sqrt{{y}}\:+\:\sqrt{{z}}\:}\:{dxdydz} \\ $$$$\: \\ $$$$\: \\ $$

Question Number 189066 Answers: 2 Comments: 0

Given f(x)+∫_0 ^1 (x+y)^2 f(y) dy=2x^2 −3x+1 find f(x).

$$\:\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} \:\mathrm{f}\left(\mathrm{y}\right)\:\mathrm{dy}=\mathrm{2x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{1} \\ $$$$\:\mathrm{find}\:\mathrm{f}\left(\mathrm{x}\right). \\ $$

Question Number 189057 Answers: 0 Comments: 6

Help! Evaluate the following integral usings Green theorem: ∮4xydx + x^2 dy Where C is the square of vertices (0,0), (0,2), (2,0) and (2,2).

$$\: \\ $$$$\:\mathrm{Help}! \\ $$$$\: \\ $$$$\:\mathrm{Evaluate}\:\:\mathrm{the}\:\:\mathrm{following}\:\:\mathrm{integral}\:\:\mathrm{usings}\:\:\mathrm{Green}\:\mathrm{theorem}: \\ $$$$\: \\ $$$$\:\oint\mathrm{4xy}{d}\mathrm{x}\:\:+\:\:\mathrm{x}^{\mathrm{2}} {d}\mathrm{y} \\ $$$$\: \\ $$$$\:\mathrm{Where}\:\:{C}\:\:\mathrm{is}\:\:\mathrm{the}\:\:\mathrm{square}\:\:\mathrm{of}\:\:\mathrm{vertices}\:\:\left(\mathrm{0},\mathrm{0}\right),\:\left(\mathrm{0},\mathrm{2}\right),\:\left(\mathrm{2},\mathrm{0}\right)\:\:\mathrm{and}\:\:\left(\mathrm{2},\mathrm{2}\right). \\ $$$$\: \\ $$

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If, y= (( Arcsin((√x) ))/( (√( x (1−x ))))) ⇒ y′ .p(x) + y .q(x)= 1 find , ∫_0 ^( 1) p(x).q(x)dx=? p , q are two pllynomils...

$$ \\ $$$$\:\:{If},\:{y}=\:\frac{\:{Arcsin}\left(\sqrt{{x}}\:\right)}{\:\sqrt{\:{x}\:\left(\mathrm{1}−{x}\:\right)}}\:\:\Rightarrow \\ $$$$\:\:\:{y}'\:.{p}\left({x}\right)\:+\:{y}\:.{q}\left({x}\right)=\:\mathrm{1} \\ $$$$ \\ $$$$\:\:\:{find}\:,\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {p}\left({x}\right).{q}\left({x}\right){dx}=? \\ $$$$\:\:\:\:{p}\:,\:{q}\:\:{are}\:{two}\:{pllynomils}... \\ $$$$ \\ $$

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evaluate ∫_0 ^π (dx/(a+bcosx )) , a > 0 and deduce that ∫_0 ^π (dx/((a+bcos x)^2 )) = ((πa)/((a^2 −b^2 )^(3/2) )) ; a^2 >b^2 and ∫_0 ^π ((cos x dx)/((a+bcos x)^2 )) = ((−πb)/((a^2 −b^2 )^(3/2) )) ; a^2 >b^2

$$\:\:\:{evaluate} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{{a}+{b}\mathrm{cos}{x}\:}\:\:\:\:\:\:,\:\:\:{a}\:>\:\mathrm{0} \\ $$$$\:\:\:{and}\:{deduce}\:{that} \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\left({a}+{b}\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\:\:=\:\:\:\frac{\pi{a}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\:\:;\:\:{a}^{\mathrm{2}} >{b}^{\mathrm{2}} \\ $$$${and}\:\:\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cos}\:{x}\:{dx}}{\left({a}+{b}\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\:\:=\:\frac{−\pi{b}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\:\:;\:\:{a}^{\mathrm{2}} >{b}^{\mathrm{2}} \\ $$

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∫_0 ^(π/4) arctan((√((1−tan^2 x)/2)))dx = ?

$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}{arctan}\left(\sqrt{\frac{\mathrm{1}−{tan}^{\mathrm{2}} {x}}{\mathrm{2}}}\right){dx}\:=\:? \\ $$

Question Number 188384 Answers: 1 Comments: 0

if ∫_0 ^∞ e^(−ax) dx = (1/a) show that ∫_0 ^∞ e^(−ax) x^n dx = ((n!)/a^(n+1) )

$$\:\:\: \\ $$$$\:\:\:\mathrm{if}\:\:\:\:\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} {dx}\:\:=\:\:\frac{\mathrm{1}}{{a}} \\ $$$$\:\:\:\:\:\:\:\:{show}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} \:{x}^{{n}} {dx}\:\:=\:\:\frac{{n}!}{{a}^{{n}+\mathrm{1}} } \\ $$

Question Number 188381 Answers: 2 Comments: 0

Advanced calculus Find the value of the following series. Ω = Σ_(n=1) ^∞ (( (−1)^( n) ζ ( n ))/(n. 2^( n) )) = ? ζ ( z ) = Σ_( n=1) ^∞ (( 1)/(n^( z) )) ; Re ( z )>1

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Advanced}\:\:\mathrm{calculus} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{Find}\:\:\mathrm{the}\:\:\mathrm{value}\:\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\:\left(−\mathrm{1}\right)^{\:{n}} \:\zeta\:\left(\:{n}\:\right)}{{n}.\:\mathrm{2}^{\:{n}} }\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\zeta\:\left(\:{z}\:\right)\:=\:\underset{\:{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{1}}{{n}^{\:{z}} \:}\:\:\:\:\:;\:\:\:\:\mathscr{R}{e}\:\left(\:{z}\:\right)>\mathrm{1} \\ $$$$ \\ $$

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