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IntegrationQuestion and Answers: Page 186

Question Number 89055 Answers: 0 Comments: 0

Question Number 89052 Answers: 0 Comments: 4

∫_0 ^(π/2) (x/(tan(x)))dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}}{{tan}\left({x}\right)}{dx} \\ $$

Question Number 89025 Answers: 1 Comments: 0

∫((3^x +4^x )/5^x )dx

$$\int\frac{\mathrm{3}^{{x}} +\mathrm{4}^{{x}} }{\mathrm{5}^{{x}} }{dx} \\ $$

Question Number 88973 Answers: 0 Comments: 0

∫((3tan(x))/(2sin^2 (x)+5cos^2 (x)+sec(x)))dx

$$\int\frac{\mathrm{3}{tan}\left({x}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left({x}\right)+\mathrm{5}{cos}^{\mathrm{2}} \left({x}\right)+{sec}\left({x}\right)}{dx} \\ $$

Question Number 88955 Answers: 0 Comments: 10

hello floor function ∫_a ^b ⌊x⌋ dx a,b∈z and b>a =∫_0 ^b ⌊x⌋ dx −∫_0 ^a ⌊x⌋ dx =((b^2 −b)/2)−((a^2 −a)/2) ....(1) now ∫_m ^k ⌊x⌋ dx when m,k∉z when m<a<b<k b=[k] and a=[m] ∫_m ^a ⌊x⌋dx +∫_a ^b ⌊x⌋dx +∫_b ^k ⌊x⌋dx =(a−m)⌊m⌋+((b^2 −b)/2)−((a^2 −a)/2)+(k−b)⌊k⌋ =(⌊m⌋−m)⌊m⌋+((⌊k⌋^2 −⌊k⌋)/2)−((⌊m⌋^2 −⌊m⌋)/2)+(k−⌊k⌋)⌊k⌋ ⌊m⌋^2 −m⌊m⌋ +(1/2)⌊k⌋^2 −(1/2)⌊k⌋−(1/2)⌊m⌋^2 −(1/2)⌊m⌋+k⌊k⌋−⌊k⌋^2 =k⌊k⌋−m⌊m⌋+(1/2)⌊m⌋^2 −(1/2)⌊k⌋^2 +(1/2)⌊m⌋−(1/2)⌊k⌋ ∴∫_m ^k ⌊x⌋dx=k⌊k⌋−m⌊m⌋+(1/2)(⌊m⌋−⌊k⌋)(⌊m⌋+⌊k⌋)+1 example ∫_(−1.5) ^(3.7) ⌊x⌋dx=(3.7)3−((−1.5)(−2))+(1/2)(−2−3)(−2+3+1) =11.1−3−5=3.1

$${hello}\: \\ $$$${floor}\:{function} \\ $$$$\int_{{a}} ^{{b}} \lfloor{x}\rfloor\:\:{dx}\:\:\:\:\:\:\:\:\:\:{a},{b}\in{z}\:\:\:{and}\:{b}>{a} \\ $$$$=\int_{\mathrm{0}} ^{{b}} \lfloor{x}\rfloor\:{dx}\:−\int_{\mathrm{0}} ^{{a}} \lfloor{x}\rfloor\:{dx}\:=\frac{{b}^{\mathrm{2}} −{b}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} −{a}}{\mathrm{2}}\:....\left(\mathrm{1}\right) \\ $$$${now} \\ $$$$\int_{{m}} ^{{k}} \lfloor{x}\rfloor\:{dx}\:\:\:{when}\:{m},{k}\notin{z}\:\:\:\:{when}\:{m}<{a}<{b}<{k} \\ $$$${b}=\left[{k}\right]\:\:\:\:{and}\:{a}=\left[{m}\right] \\ $$$$\int_{{m}} ^{{a}} \lfloor{x}\rfloor{dx}\:+\int_{{a}} ^{{b}} \lfloor{x}\rfloor{dx}\:+\int_{{b}} ^{{k}} \lfloor{x}\rfloor{dx} \\ $$$$=\left({a}−{m}\right)\lfloor{m}\rfloor+\frac{{b}^{\mathrm{2}} −{b}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} −{a}}{\mathrm{2}}+\left({k}−{b}\right)\lfloor{k}\rfloor \\ $$$$=\left(\lfloor{m}\rfloor−{m}\right)\lfloor{m}\rfloor+\frac{\lfloor{k}\rfloor^{\mathrm{2}} −\lfloor{k}\rfloor}{\mathrm{2}}−\frac{\lfloor{m}\rfloor^{\mathrm{2}} −\lfloor{m}\rfloor}{\mathrm{2}}+\left({k}−\lfloor{k}\rfloor\right)\lfloor{k}\rfloor \\ $$$$\lfloor{m}\rfloor^{\mathrm{2}} −{m}\lfloor{m}\rfloor\:+\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor−\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor+{k}\lfloor{k}\rfloor−\lfloor{k}\rfloor^{\mathrm{2}} \\ $$$$={k}\lfloor{k}\rfloor−{m}\lfloor{m}\rfloor+\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor−\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor \\ $$$$\therefore\int_{{m}} ^{{k}} \lfloor{x}\rfloor{dx}={k}\lfloor{k}\rfloor−{m}\lfloor{m}\rfloor+\frac{\mathrm{1}}{\mathrm{2}}\left(\lfloor{m}\rfloor−\lfloor{k}\rfloor\right)\left(\lfloor{m}\rfloor+\lfloor{k}\rfloor\right)+\mathrm{1} \\ $$$${example} \\ $$$$\int_{−\mathrm{1}.\mathrm{5}} ^{\mathrm{3}.\mathrm{7}} \lfloor{x}\rfloor{dx}=\left(\mathrm{3}.\mathrm{7}\right)\mathrm{3}−\left(\left(−\mathrm{1}.\mathrm{5}\right)\left(−\mathrm{2}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{2}−\mathrm{3}\right)\left(−\mathrm{2}+\mathrm{3}+\mathrm{1}\right) \\ $$$$=\mathrm{11}.\mathrm{1}−\mathrm{3}−\mathrm{5}=\mathrm{3}.\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Question Number 88951 Answers: 1 Comments: 0

∫_a ^b ⌈x⌉ dx=? a,b∈R ⌈..⌉ is ceil

$$\int_{{a}} ^{{b}} \lceil{x}\rceil\:{dx}=? \\ $$$${a},{b}\in{R}\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\lceil..\rceil\:{is}\:{ceil}\: \\ $$

Question Number 88930 Answers: 0 Comments: 2

find A_λ =∫_0 ^∞ ((cos(λx))/((x^2 −x+1)^2 ))dx with λ>0 2)find the value of ∫_0 ^∞ ((cos(3x))/((x^2 −x+1)^2 ))dx

$${find}\:{A}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\lambda{x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:{with}\:\lambda>\mathrm{0} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 88929 Answers: 0 Comments: 1

cakculate ∫_0 ^∞ ((arctan(ch(x)))/(4+x^2 ))dx

$${cakculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({ch}\left({x}\right)\right)}{\mathrm{4}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 88928 Answers: 0 Comments: 4

calculate ∫_0 ^∞ (dx/((x^4 +x^2 +3)^2 ))

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} \:\:+{x}^{\mathrm{2}} \:\:+\mathrm{3}\right)^{\mathrm{2}} } \\ $$

Question Number 88927 Answers: 0 Comments: 0

calculate ∫_0 ^∞ ((sin(∣arctanx∣))/(x^2 +1))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mid{arctanx}\mid\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$

Question Number 88902 Answers: 1 Comments: 4

∫_(1/e) ^e ln∣x∣ dx

$$\int_{\frac{\mathrm{1}}{{e}}} ^{{e}} {ln}\mid{x}\mid\:{dx} \\ $$

Question Number 88863 Answers: 0 Comments: 1

∫((8cos^3 (x))/(8+sin^3 2x))dx

$$\int\frac{\mathrm{8}{cos}^{\mathrm{3}} \left({x}\right)}{\mathrm{8}+{sin}^{\mathrm{3}} \mathrm{2}{x}}{dx} \\ $$$$ \\ $$

Question Number 88859 Answers: 0 Comments: 1

∫ e^(ax) cos bx dx ∫x^2 e^(2x) ln3x^2 dx

$$\int\:\boldsymbol{{e}}^{\boldsymbol{{ax}}} \mathrm{cos}\:\boldsymbol{{bx}}\:\boldsymbol{{dx}} \\ $$$$\int\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}} \boldsymbol{{ln}}\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} \:\boldsymbol{{dx}} \\ $$

Question Number 88852 Answers: 1 Comments: 0

prove that ∫_0 ^n ⌈x⌉dx= ((n(n+1))/2) and ∫_0 ^n ⌊x⌋dx=((n(n−1))/2) when ⌊..⌋ is floor and ⌈..⌉ is ceil

$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{{n}} \lceil{x}\rceil{dx}=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:{and}\:\int_{\mathrm{0}} ^{{n}} \lfloor{x}\rfloor{dx}=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${when}\:\lfloor..\rfloor\:{is}\:{floor}\:{and}\:\lceil..\rceil\:{is}\:{ceil} \\ $$

Question Number 88789 Answers: 0 Comments: 7

∫∫ln(x+1) dx dy

$$\int\int{ln}\left({x}+\mathrm{1}\right)\:{dx}\:{dy} \\ $$

Question Number 88710 Answers: 2 Comments: 1

∫_(−(√3) ) ^(√3) ∫_1 ^(√(4−x^2 )) (x^2 +y^2 )^(3/2) dydx

$$\underset{−\sqrt{\mathrm{3}}\:} {\overset{\sqrt{\mathrm{3}}} {\int}}\underset{\mathrm{1}} {\overset{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} {\int}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} \:{dydx} \\ $$

Question Number 88616 Answers: 1 Comments: 9

Question Number 88606 Answers: 2 Comments: 2

Question Number 88603 Answers: 1 Comments: 0

Question Number 88586 Answers: 1 Comments: 0

∫((√(cos(2x)+3))/(cos(x)))dx

$$\int\frac{\sqrt{{cos}\left(\mathrm{2}{x}\right)+\mathrm{3}}}{{cos}\left({x}\right)}{dx} \\ $$

Question Number 88569 Answers: 1 Comments: 1

Question Number 88555 Answers: 1 Comments: 0

slove ⌈(x/a)⌉<a when a>1 ⌈...⌉ is ceil function

$${slove}\: \\ $$$$\lceil\frac{{x}}{{a}}\rceil<{a}\:\:\: \\ $$$${when}\:{a}>\mathrm{1} \\ $$$$\lceil...\rceil\:{is}\:{ceil}\:{function} \\ $$

Question Number 88547 Answers: 1 Comments: 0

prove for (0<a<2) ∫_0 ^( ∞) ((x^(a−1) dx)/(1+x+x^2 )) = ((2π)/(√3))cos (((2πa+π)/6))cosec πa .

$${prove}\:{for}\:\left(\mathrm{0}<{a}<\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{{a}−\mathrm{1}} {dx}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\mathrm{cos}\:\left(\frac{\mathrm{2}\pi{a}+\pi}{\mathrm{6}}\right)\mathrm{cosec}\:\pi{a}\:. \\ $$

Question Number 88490 Answers: 0 Comments: 4

∫_1 ^∞ (x^4 /4^x )dx=?

$$\int_{\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{4}} }{\mathrm{4}^{{x}} }{dx}=? \\ $$

Question Number 88462 Answers: 0 Comments: 3

Question Number 88438 Answers: 2 Comments: 0

∫((x^5 +1)/(x^5 −1))dx

$$\int\frac{{x}^{\mathrm{5}} +\mathrm{1}}{{x}^{\mathrm{5}} −\mathrm{1}}{dx} \\ $$

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