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Category: Integration

Given-that-F-x-0-x-t-2-t-2-1-dt-Show-that-F-x-is-an-increasing-function-

Question Number 146455 by physicstutes last updated on 13/Jul/21 $$\mathrm{Given}\:\mathrm{that}\:\:{F}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{{x}} \frac{{t}^{\mathrm{2}} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}\: \\ $$$$\:\mathrm{Show}\:\mathrm{that}\:{F}\left({x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{increasing}\:\mathrm{function} \\ $$ Answered by gsk2684 last updated on 13/Jul/21…

1-Integrate-F-x-y-x-2-over-the-region-bounded-by-y-x-2-x-2-and-x-1-2-Integrate-G-x-y-x-2-y-2-over-the-region-bounded-by-the-triangle-x-y-y-

Question Number 80914 by TawaTawa last updated on 07/Feb/20 $$\left(\mathrm{1}\right) \\ $$$$\mathrm{Integrate}\:\:\mathrm{F}\left(\mathrm{x},\:\mathrm{y}\right)\:\:=\:\:\mathrm{x}^{\mathrm{2}} \:\:\:\mathrm{over}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\:\:\mathrm{y}\:\:=\:\:\mathrm{x}^{\mathrm{2}} , \\ $$$$\mathrm{x}\:\:=\:\:\mathrm{2}\:\:\mathrm{and}\:\mathrm{x}\:\:=\:\:\mathrm{1} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{Integrate}\:\:\:\:\mathrm{G}\left(\mathrm{x},\:\mathrm{y}\right)\:\:=\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:\:\:\:\mathrm{over}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\: \\…

n-1-n-ln-2n-1-2n-1-1-

Question Number 146429 by qaz last updated on 13/Jul/21 $$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{n}\centerdot\mathrm{ln}\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{2n}−\mathrm{1}}−\mathrm{1}\right)=? \\ $$ Commented by qaz last updated on 14/Jul/21 $$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{n}\centerdot\mathrm{ln}\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{2n}−\mathrm{1}}−\mathrm{1}\right) \\…

find-1-x-n-1-dx-for-n-N-

Question Number 146361 by gsk2684 last updated on 13/Jul/21 $${find}\:\int\frac{\mathrm{1}}{{x}^{{n}} +\mathrm{1}}{dx}\:{for}\:{n}\in{N} \\ $$ Answered by mathmax by abdo last updated on 13/Jul/21 $$\mathrm{z}^{\mathrm{n}} \:+\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{z}^{\mathrm{n}} \:=\mathrm{e}^{\left(\mathrm{2k}+\mathrm{1}\right)\mathrm{i}\pi}…