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Given ∫_0 ^1 f(x) dx = (((2018)),(( 0)) ) + (1/2) (((2018)),(( 1)) ) + (1/3) (((2018)),(( 2)) ) + ... + (1/(2019)) (((2018)),((2018)) ) ∫_0 ^1 g(x) dx = (((2018)),(( 0)) ) − (1/2) (((2018)),(( 1)) ) + (1/3) (((2018)),(( 2)) ) − ... + (1/(2019)) (((2018)),((2018)) ) h(x) is an odd function Then what is the value of ∫_(−3) ^( 3) f(x).g(x).h(x) dx ?

$$\mathrm{Given} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{f}\left({x}\right)\:{dx}\:=\:\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{0}}\end{pmatrix}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{1}}\end{pmatrix}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{2}}\end{pmatrix}\:+\:...\:+\:\frac{\mathrm{1}}{\mathrm{2019}}\begin{pmatrix}{\mathrm{2018}}\\{\mathrm{2018}}\end{pmatrix} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{g}\left({x}\right)\:{dx}\:=\:\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{0}}\end{pmatrix}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{1}}\end{pmatrix}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{2}}\end{pmatrix}\:−\:...\:+\:\frac{\mathrm{1}}{\mathrm{2019}}\begin{pmatrix}{\mathrm{2018}}\\{\mathrm{2018}}\end{pmatrix} \\ $$$${h}\left({x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function} \\ $$$$\mathrm{Then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\int_{−\mathrm{3}} ^{\:\mathrm{3}} \:{f}\left({x}\right).{g}\left({x}\right).{h}\left({x}\right)\:{dx}\:? \\ $$

Question Number 31141 Answers: 1 Comments: 0

using the limit defination find the area of f(x)= cos(x) [0,π/2]

$$\boldsymbol{{using}}\:\boldsymbol{{the}}\:\boldsymbol{{limit}}\:\boldsymbol{{defination}} \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{area}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\:\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\:\:\left[\mathrm{0},\pi/\mathrm{2}\right] \\ $$

Question Number 30936 Answers: 0 Comments: 0

find ∫^a _0 ((sinx)/(√(1+x^2 )))dx .

$${find}\:\:\:\underset{\mathrm{0}} {\int}^{{a}} \:\:\frac{{sinx}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}\:. \\ $$

Question Number 30858 Answers: 2 Comments: 0

∫_0 ^1 x∣x−4∣dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}\mid\mathrm{x}−\mathrm{4}\mid\mathrm{dx} \\ $$

Question Number 30855 Answers: 1 Comments: 0

∫((cosec^2 (x))/(√(cosecx+cotx)))dx

$$\int\frac{\mathrm{cosec}^{\mathrm{2}} \left(\mathrm{x}\right)}{\sqrt{\mathrm{cosecx}+\mathrm{cotx}}}\mathrm{dx} \\ $$

Question Number 30798 Answers: 0 Comments: 1

find ∫_0 ^1 ((ln(1−x^2 ))/x^2 )dx

$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 30796 Answers: 0 Comments: 0

find ∫_0 ^∞ e^(−(x^2 +(1/x^2 ))) dx.

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

Question Number 30777 Answers: 0 Comments: 0

find interms of n A_n = ∫_0 ^∞ ((ln(x))/((1+x^ )^n )) dx with n from N and n≥3 .

$${find}\:{interms}\:{of}\:{n}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}^{} \right)^{{n}} }\:{dx}\:{with}\:{n}\:{from} \\ $$$${N}\:{and}\:{n}\geqslant\mathrm{3}\:. \\ $$

Question Number 30776 Answers: 1 Comments: 1

find ∫_0 ^1 ((xdx)/((1+x^2 )(√(1−x^4 )))) .

$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{xdx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:. \\ $$

Question Number 30775 Answers: 1 Comments: 1

letα ∈]0,π[ calculate ∫_0 ^(π/2) (dx/(2(cosα +chx))) .

$$\left.{let}\alpha\:\in\right]\mathrm{0},\pi\left[\:\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{2}\left({cos}\alpha\:+{chx}\right)}\:.\right. \\ $$

Question Number 30774 Answers: 1 Comments: 1

find f(t)=∫_0 ^1 ln(1+tx^2 )dx with t>0

$${find}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}+{tx}^{\mathrm{2}} \right){dx}\:\:{with}\:{t}>\mathrm{0} \\ $$

Question Number 30773 Answers: 0 Comments: 1

let a>0 calculate ∫_0 ^∞ (dx/((x^2 +a^2 )^2 )) 2) calculate ∫_(−∞) ^(+∞) (x^2 /((x^2 +a^2 )^2 ))dx.

$${let}\:{a}>\mathrm{0}\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}. \\ $$

Question Number 30769 Answers: 0 Comments: 1

find the value of I= ∫_0 ^1 (dx/((x+1)^2 (√(x^2 +2x +2)))) .

$${find}\:{the}\:{value}\:{of}\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}}}\:. \\ $$

Question Number 30767 Answers: 0 Comments: 1

calculate A_n = ∫_0 ^1 ((1−(√x))/(1−^n (√x)))dx.

$${calculate}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}−^{{n}} \sqrt{{x}}}{dx}. \\ $$

Question Number 30766 Answers: 0 Comments: 0

find ∫_0 ^1 (x/(√(x^4 +x^2 +1)))dx

$${find}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{{x}}{\sqrt{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}}}{dx} \\ $$

Question Number 30765 Answers: 0 Comments: 1

find ∫_0 ^1 (dx/(√(x^2 +x+1))) .

$${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}}\:. \\ $$

Question Number 30764 Answers: 0 Comments: 1

let I_n = ∫_0 ^(1/2) (1−2t)^n e^(−t) dt with n integr not 0 1) prove that ∀t∈[0,(1/2)] (1/(√e))(1−2t)^n ≤ (1−2t)^n e^(−t) ≤(1−2t)^n then find lim_(n→∞ ) I_n 2) prove that I_(n+1 ) =1−2(n+1)I_n 3) calculate I_1 ,I_2 , and I_3 .

$${let}\:\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \:{e}^{−{t}} {dt}\:\:{with}\:{n}\:{integr}\:{not}\:\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\forall{t}\in\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$\frac{\mathrm{1}}{\sqrt{{e}}}\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \leqslant\:\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \:{e}^{−{t}} \:\leqslant\left(\mathrm{1}−\mathrm{2}{t}\right)^{{n}} \:{then}\:{find}\:{lim}_{{n}\rightarrow\infty\:} {I}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:{I}_{{n}+\mathrm{1}\:} =\mathrm{1}−\mathrm{2}\left({n}+\mathrm{1}\right){I}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{I}_{\mathrm{1}} ,{I}_{\mathrm{2}} ,\:{and}\:{I}_{\mathrm{3}} . \\ $$

Question Number 30761 Answers: 0 Comments: 1

find ∫_0 ^∞ x^(2n+1) e^(−x^2 ) dx with n from N.

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:\:\:{with}\:{n}\:{from}\:{N}. \\ $$

Question Number 30760 Answers: 0 Comments: 1

find I_n = ∫_0 ^1 (lnx)^n dx with n fromN

$${find}\:\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left({lnx}\right)^{{n}} \:{dx}\:\:{with}\:{n}\:{fromN} \\ $$

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