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Category: Integration

Help-me-Use-double-integral-to-find-the-area-of-the-region-bounded-by-the-following-curves-given-in-the-plane-shown-below-y-2-4x-and-x-2-4y-

Question Number 179181 by neinhaltsieger369 last updated on 26/Oct/22 $$\:\mathrm{Help}-\mathrm{me}! \\ $$$$\: \\ $$$$\:\mathrm{Use}\:\mathrm{double}\:\mathrm{integral}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{following}\:\mathrm{curves}\: \\ $$$$\:\mathrm{given}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{shown}\:\mathrm{below}: \\ $$$$\: \\ $$$$\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{4x}\:\mathrm{and}\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:=\:\mathrm{4y} \\…

solve-this-2-sinx-cosx-2-3sinx-sin-2x-dx-

Question Number 48104 by wasim last updated on 19/Nov/18 $$\mathrm{solve}\:\mathrm{this}\:\: \\ $$$$\int\left(\mathrm{2}\:\mathrm{sinx}+\mathrm{cosx}\right)/\left(\mathrm{2}+\mathrm{3sinx}+\mathrm{sin}^{\mathrm{2x}} \right)\:\mathrm{dx} \\ $$ Answered by MJS last updated on 19/Nov/18 $$\mathrm{Weierstrass}−\mathrm{substitution} \\ $$$${t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\Rightarrow\:{x}=\mathrm{2arctan}\:{t};\:{dx}=\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}}…

find-dx-x-1-x-2-1-x-1-x-2-1-

Question Number 113628 by mathmax by abdo last updated on 14/Sep/20 $$\mathrm{find}\:\int\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}+\left(\mathrm{x}−\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}} \\ $$ Answered by MJS_new last updated on 16/Sep/20 $$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\left({x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}}…

Question-48078

Question Number 48078 by cesar.marval.larez@gmail.com last updated on 19/Nov/18 Answered by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18 $$\left.\mathrm{21}\right)\int{e}^{\mathrm{2}−{x}} {dx} \\ $$$${t}=\mathrm{2}−{x}\:\:\:{dt}=−{dx} \\ $$$$\int{e}^{{t}} ×−{dt} \\ $$$$=\left(−\mathrm{1}\right){e}^{{t}}…