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IntegrationQuestion and Answers: Page 186

Question Number 89362    Answers: 0   Comments: 1

∫(1/((√x)((√x)+1)^3 ))

$$\int\frac{\mathrm{1}}{\sqrt{\mathrm{x}}\left(\sqrt{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Question Number 89286    Answers: 2   Comments: 0

show that ∫_0 ^1 (((x^2 +1)ln(1+x))/(x^4 −x^2 +1))dx=(π/6)ln(2+(√3))

$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right){ln}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\frac{\pi}{\mathrm{6}}{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$

Question Number 89273    Answers: 1   Comments: 0

Evaluate ∫_0 ^1 (x^2 /(√(1+x^3 )))dx and given that I_(n ) =∫_0 ^1 x^n (1+x^3 )^(−(1/2)) dx show that (2n−1)I_n =2(√2)−2(n−1) for n≥3. Hence evaluate I_8 , I_7 and I_6

$${Evaluate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }}{dx}\:{and}\:{given}\:{that}\:{I}_{{n}\:} =\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$${show}\:{that}\:\left(\mathrm{2}{n}−\mathrm{1}\right){I}_{{n}} =\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}\left({n}−\mathrm{1}\right)\:{for}\:{n}\geqslant\mathrm{3}. \\ $$$${Hence}\:{evaluate}\:{I}_{\mathrm{8}} ,\:{I}_{\mathrm{7}} \:{and}\:{I}_{\mathrm{6}} \\ $$

Question Number 89311    Answers: 0   Comments: 0

Show that ∫_( 0) ^( 1) {∫_( 0) ^( 1) ((x−y)/((x+y)^2 ))dy}dx=∫_( 0) ^( 1) {∫_( 0) ^( 1) ((x−y)/((x+y)^2 ))dx}dy

$$\:\:{Show}\:{that} \\ $$$$\underset{\:\:\:\mathrm{0}} {\overset{\:\:\:\:\:\:\:\mathrm{1}} {\int}}\left\{\underset{\:\:\:\:\mathrm{0}} {\overset{\:\:\:\mathrm{1}} {\int}}\frac{{x}−{y}}{\left({x}+{y}\right)^{\mathrm{2}} }{dy}\right\}{dx}=\underset{\:\:\mathrm{0}} {\overset{\:\:\:\:\:\:\:\mathrm{1}} {\int}}\left\{\underset{\:\:\:\mathrm{0}} {\overset{\:\:\:\:\:\:\mathrm{1}} {\int}}\frac{{x}−{y}}{\left({x}+{y}\right)^{\mathrm{2}} }{dx}\right\}{dy} \\ $$$$ \\ $$

Question Number 89237    Answers: 0   Comments: 1

∫_2 ^1 (x+1)((√(x+3)))

$$\int_{\mathrm{2}} ^{\mathrm{1}} \left(\mathrm{x}+\mathrm{1}\right)\left(\sqrt{\left.\mathrm{x}+\mathrm{3}\right)}\right. \\ $$

Question Number 89240    Answers: 0   Comments: 0

If I_n =∫_0 ^π e^x sin^n xdx, show that (n^2 +1)I_n =n(n−1)I_(n−2)

$${If}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\pi} {e}^{{x}} {sin}^{{n}} {xdx},\:{show}\:{that}\: \\ $$$$\left({n}^{\mathrm{2}} +\mathrm{1}\right){I}_{{n}} ={n}\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} \\ $$

Question Number 89226    Answers: 0   Comments: 1

∫(((x^3 +2)/x^3 ))(√(x−(1/x^2 )))

$$\int\left(\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{2}}{\mathrm{x}^{\mathrm{3}} }\right)\sqrt{\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \\ $$

Question Number 89213    Answers: 1   Comments: 0

∫((√(tan x + 1))/(cos^2 x))

$$\int\frac{\sqrt{\mathrm{tan}\:\mathrm{x}\:+\:\mathrm{1}}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}} \\ $$

Question Number 89228    Answers: 0   Comments: 1

∫_4 ^5 x^2 (√(x−4))

$$\underset{\mathrm{4}} {\overset{\mathrm{5}} {\int}}\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}−\mathrm{4}} \\ $$

Question Number 89146    Answers: 6   Comments: 4

1)∫x(√((x−2)/(x+1))) dx 2)∫(1/((x+1)(√(x^2 +x+1))))dx 3)∫(√(−x^2 +4x+10)) dx

$$\left.\mathrm{1}\right)\int{x}\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}\:{dx} \\ $$$$\left.\mathrm{2}\right)\int\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx} \\ $$$$\left.\mathrm{3}\right)\int\sqrt{−{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}\:{dx} \\ $$

Question Number 89123    Answers: 0   Comments: 4

∫ ((cos 2x)/(sec x−cos^2 x)) dx ?

$$\int\:\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{sec}\:{x}−\mathrm{cos}\:^{\mathrm{2}} {x}}\:{dx}\:? \\ $$

Question Number 89161    Answers: 1   Comments: 2

∫_0 ^(π/2) log(sin(x))dx

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

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

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

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