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IntegrationQuestion and Answers: Page 151

Question Number 109546 Answers: 1 Comments: 2

If f(x) continue in [ 1,30] and ∫_6 ^(30) f(x)dx = 30, then ∫_1 ^9 f(3y+3)dy = __

$${If}\:{f}\left({x}\right)\:{continue}\:{in}\:\left[\:\mathrm{1},\mathrm{30}\right]\:{and}\: \\ $$$$\underset{\mathrm{6}} {\overset{\mathrm{30}} {\int}}{f}\left({x}\right){dx}\:=\:\mathrm{30},\:{then}\:\underset{\mathrm{1}} {\overset{\mathrm{9}} {\int}}{f}\left(\mathrm{3}{y}+\mathrm{3}\right){dy}\:=\:\_\_ \\ $$

Question Number 109506 Answers: 0 Comments: 0

Question Number 109509 Answers: 4 Comments: 0

((bemath)/(Σ_(i=cooll) ^(nice) (joss)_i )) ∫ ((x^2 dx)/( (√(x^2 +25))))

$$\:\:\frac{{bemath}}{\underset{{i}={cooll}} {\overset{{nice}} {\sum}}\left({joss}\right)_{{i}} }\: \\ $$$$ \\ $$$$\int\:\frac{{x}^{\mathrm{2}} \:{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}} \\ $$

Question Number 109472 Answers: 4 Comments: 0

Question Number 109459 Answers: 0 Comments: 0

Question Number 109457 Answers: 3 Comments: 0

Question Number 109435 Answers: 1 Comments: 0

Question Number 109378 Answers: 4 Comments: 0

Question Number 109366 Answers: 0 Comments: 1

Question Number 109343 Answers: 2 Comments: 1

Question Number 109342 Answers: 0 Comments: 3

Question Number 109220 Answers: 0 Comments: 0

calculate I_n =∫_0 ^(2π) ((cos(nx))/(cosx +sinx))dx (n→natural)

$$\mathrm{calculate}\:\:\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\mathrm{cosx}\:+\mathrm{sinx}}\mathrm{dx}\:\:\left(\mathrm{n}\rightarrow\mathrm{natural}\right) \\ $$

Question Number 109219 Answers: 0 Comments: 0

let f(x) =((sin(αx))/(sinx)) , 2π periodi even developp f at fourier serie

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{sin}\left(\alpha\mathrm{x}\right)}{\mathrm{sinx}}\:\:\:\:\:,\:\mathrm{2}\pi\:\mathrm{periodi}\:\mathrm{even} \\ $$$$\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$

Question Number 109218 Answers: 0 Comments: 0

calculate ∫_0 ^∞ ((arctan(2+2t^2 ))/(1+t^2 ))dt

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{2}+\mathrm{2t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$

Question Number 109215 Answers: 0 Comments: 1

calculate ∫_0 ^1 ((ln(1+(√(1+x^2 ))))/(√(1+x^2 )))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$

Question Number 109214 Answers: 1 Comments: 0

calculateA_n = ∫_0 ^∞ (dx/((x^2 +n)(x^2 +2n))) with n integr natural≥1

$$\mathrm{calculateA}_{\mathrm{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{n}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{2n}\right)}\:\:\mathrm{with}\:\mathrm{n}\:\mathrm{integr}\:\mathrm{natural}\geqslant\mathrm{1} \\ $$

Question Number 109212 Answers: 0 Comments: 0

calculate ∫_0 ^π ((sin(nx))/(cosx))dx with n integr

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{sin}\left(\mathrm{nx}\right)}{\mathrm{cosx}}\mathrm{dx}\:\:\mathrm{with}\:\mathrm{n}\:\mathrm{integr} \\ $$

Question Number 109136 Answers: 1 Comments: 0

∫_0 ^(1/2) ((ln(1-t)ln(t))/t) dt I′m about to give up

$$\int_{\mathrm{0}} ^{\mathrm{1}/\mathrm{2}} \frac{{ln}\left(\mathrm{1}-{t}\right){ln}\left({t}\right)}{{t}}\:{dt} \\ $$$${I}'{m}\:{about}\:{to}\:{give}\:{up} \\ $$

Question Number 109129 Answers: 1 Comments: 0

∫_0 ^(π/2) ((ln (cos x)ln (sin x))/(tan x)) dx

$$\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{tan}\:{x}}\:{dx} \\ $$

Question Number 109101 Answers: 3 Comments: 0

Given a function f(x+3)=f(x) for ∀x∈R. If ∫_(−3) ^6 f(x)dx = −6 then ∫_3 ^9 f(x) dx = ?

$$\:{Given}\:{a}\:{function}\:{f}\left({x}+\mathrm{3}\right)={f}\left({x}\right) \\ $$$${for}\:\forall{x}\in\mathbb{R}.\:{If}\:\underset{−\mathrm{3}} {\overset{\mathrm{6}} {\int}}{f}\left({x}\right){dx}\:=\:−\mathrm{6}\: \\ $$$${then}\:\underset{\mathrm{3}} {\overset{\mathrm{9}} {\int}}{f}\left({x}\right)\:{dx}\:=\:? \\ $$

Question Number 109097 Answers: 1 Comments: 0

((♭o♭hans)/(∼∼∼∼∼)) ∫_1 ^2 x sec^(−1) (x)dx=?

$$\:\:\frac{\boldsymbol{\flat{o}\flat{hans}}}{\sim\sim\sim\sim\sim} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}{x}\:\mathrm{sec}^{−\mathrm{1}} \left({x}\right){dx}=? \\ $$

Question Number 109086 Answers: 1 Comments: 0

♭_→ o_→ ♭h_⊸ ans_⊸ (1) (x^2 e^(−(y/x)) +y^2 ) dx = xy dy (2)(((f(x))/x))′ = x^2 e^(−x^2 ) ; f(1) = (1/e) g(x) = (4/e^4 )∫_1 ^x e^t^2 f(t) dt . find f(2)−g(2)

$$\:\:\:\underset{\rightarrow} {\flat}\underset{\rightarrow} {{o}}\flat\underset{\multimap} {{h}an}\underset{\multimap} {{s}} \\ $$$$\left(\mathrm{1}\right)\:\left({x}^{\mathrm{2}} {e}^{−\frac{{y}}{{x}}} +{y}^{\mathrm{2}} \right)\:{dx}\:=\:{xy}\:{dy}\: \\ $$$$\left(\mathrm{2}\right)\left(\frac{{f}\left({x}\right)}{{x}}\right)'\:=\:{x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } \:;\:{f}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{1}}{{e}}\: \\ $$$$\:\:\:\:\:\:\:{g}\left({x}\right)\:=\:\frac{\mathrm{4}}{{e}^{\mathrm{4}} }\underset{\mathrm{1}} {\overset{{x}} {\int}}{e}^{{t}^{\mathrm{2}} } \:{f}\left({t}\right)\:{dt}\:.\:{find}\:{f}\left(\mathrm{2}\right)−{g}\left(\mathrm{2}\right) \\ $$

Question Number 109047 Answers: 1 Comments: 0

∫_(−2) ^∞ (x+2)^5 e^(−(x+2)) dx ∫_0 ^1 ((tan(x))/x)dx

$$\int_{−\mathrm{2}} ^{\infty} \left({x}+\mathrm{2}\right)^{\mathrm{5}} {e}^{−\left({x}+\mathrm{2}\right)} {dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}\left({x}\right)}{{x}}{dx} \\ $$

Question Number 108998 Answers: 0 Comments: 0

Question Number 108990 Answers: 2 Comments: 0

Question Number 108921 Answers: 3 Comments: 0

a. ∫((sin^3 4x)/(cos^8 4x))dx b. ∫_(−(π/2)) ^(π/2) (x^2 e^(cosx) −2x)sinxdx

$$\mathrm{a}.\:\:\int\frac{\mathrm{sin}^{\mathrm{3}} \mathrm{4x}}{\mathrm{cos}^{\mathrm{8}} \mathrm{4x}}\mathrm{dx} \\ $$$$\mathrm{b}.\:\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{cosx}} −\mathrm{2x}\right)\mathrm{sinxdx} \\ $$

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