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Category: Integration

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-2018-0-1-2-2

Question Number 31145 by Joel578 last updated on 03/Mar/18 $$\mathrm{Given} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{f}\left({x}\right)\:{dx}\:=\:\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{0}}\end{pmatrix}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{1}}\end{pmatrix}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{2}}\end{pmatrix}\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{2019}}\begin{pmatrix}{\mathrm{2018}}\\{\mathrm{2018}}\end{pmatrix} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{g}\left({x}\right)\:{dx}\:=\:\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{0}}\end{pmatrix}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{1}}\end{pmatrix}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{2}}\end{pmatrix}\:−\:…\:+\:\frac{\mathrm{1}}{\mathrm{2019}}\begin{pmatrix}{\mathrm{2018}}\\{\mathrm{2018}}\end{pmatrix} \\ $$$${h}\left({x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function} \\ $$$$\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}\:? \\…

0-1-log-1-x-7-1-x-7-dx-

Question Number 162219 by mathlove last updated on 27/Dec/21 $$\Omega=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{log}\left(\mathrm{1}+{x}^{\mathrm{7}} \right)}{\mathrm{1}+{x}^{\mathrm{7}} }{dx}=? \\ $$ Answered by amin96 last updated on 27/Dec/21 $$\boldsymbol{{x}}^{\mathrm{7}} =−\boldsymbol{{t}}\:\:\:\:\frac{\boldsymbol{{dt}}}{\boldsymbol{{dx}}}=−\mathrm{7}\boldsymbol{{x}}^{\mathrm{6}}…

using-the-limit-defination-find-the-area-of-f-x-cos-x-0-pi-2-

Question Number 31141 by Cheyboy last updated on 03/Mar/18 $$\boldsymbol{{using}}\:\boldsymbol{{the}}\:\boldsymbol{{limit}}\:\boldsymbol{{defination}} \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{area}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\:\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\:\:\left[\mathrm{0},\pi/\mathrm{2}\right] \\ $$ Answered by Joel578 last updated on 03/Mar/18 $${A}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…

Evaluate-log-x-a-x-dx-

Question Number 96672 by 175 last updated on 03/Jun/20 $${Evaluate}\:: \\ $$$$\int\:\frac{{log}_{{x}} {a}}{{x}}\:{dx} \\ $$ Commented by bemath last updated on 03/Jun/20 $$\int\:\frac{\mathrm{ln}\left(\mathrm{a}\right)}{\mathrm{x}\:\mathrm{ln}\left(\mathrm{x}\right)}\:\mathrm{dx}\:=\:\mathrm{ln}\left(\mathrm{a}\right)\:\int\:\frac{\mathrm{du}}{\mathrm{u}} \\ $$$$=\:\mathrm{ln}\left(\mathrm{a}\right)\:\mathrm{ln}\left(\mathrm{u}\right)\:+\:\mathrm{c}\:…

xcos-x-sin-x-x-2-sin-2-x-dx-

Question Number 96652 by bobhans last updated on 03/Jun/20 $$\int\:\frac{{x}\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} +\mathrm{sin}\:^{\mathrm{2}} {x}}\:{dx}\: \\ $$ Answered by bemath last updated on 03/Jun/20 $$\int\:\frac{\left(\frac{\mathrm{cos}\:{x}}{{x}}−\frac{\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} }\right)\:{dx}}{\mathrm{1}+\left(\frac{\mathrm{sin}\:{x}}{{x}}\right)^{\mathrm{2}} }\:=\mathrm{I} \\…

prove-that-0-x-e-t-2-dt-pi-2-e-x-2-pi-0-e-x-2-t-2-1-t-2-dt-with-x-gt-0-

Question Number 31105 by abdo imad last updated on 02/Mar/18 $${prove}\:{that}\:\int_{\mathrm{0}} ^{{x}} \:\:\:{e}^{−{t}^{\mathrm{2}} } {dt}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:−\frac{{e}^{−{x}^{\mathrm{2}} } }{\:\sqrt{\pi}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:{with}\:{x}>\mathrm{0} \\ $$…

prove-that-0-e-x-2-lim-n-0-dx-1-x-2-n-2-prove-that-1-pi-lim-n-1-3-5-2n-3-2-4-6-2n-2-n-wallis-formula-

Question Number 31106 by abdo imad last updated on 02/Mar/18 $${prove}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}^{\mathrm{2}} } ={lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\:. \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\:\frac{\mathrm{1}}{\:\sqrt{\pi}}\:={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}−\mathrm{3}\right)}{\mathrm{2}.\mathrm{4}.\mathrm{6}….\left(\mathrm{2}{n}−\mathrm{2}\right)}\:\sqrt{{n}} \\ $$$$\left({wallis}\:{formula}\right).…