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Category: Integration

Question-47915

Question Number 47915 by behi83417@gmail.com last updated on 16/Nov/18 Answered by MJS last updated on 16/Nov/18 $${f}\left({x}\right)\:\mathrm{is}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{a}\:\mathrm{parabola}.\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we} \\ $$$$\mathrm{can}\:\mathrm{find}\:{f}^{−\mathrm{1}} \left({x}\right)\:\mathrm{but}\:\mathrm{of}\:\mathrm{course}\:\mathrm{we}\:\mathrm{can}\:\mathrm{approximately} \\ $$$$\mathrm{calculate}\:\mathrm{the}\:\mathrm{area}.\:\mathrm{area}\:\mathrm{between}\:{f}\:\mathrm{and}\:{f}^{−\mathrm{1}} = \\ $$$$=\mathrm{2}×\mathrm{area}\:\mathrm{between}\:{f}\:\mathrm{and}\:{y}={x}…

let-f-x-x-1-x-and-g-x-x-1-x-find-f-x-g-x-dx-and-f-x-dx-g-x-dx-

Question Number 47851 by maxmathsup by imad last updated on 15/Nov/18 $${let}\:\:{f}\left({x}\right)={x}+\mathrm{1}+\sqrt{{x}}\:{and}\:{g}\left({x}\right)={x}+\mathrm{1}−\sqrt{{x}} \\ $$$${find}\:\int\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}{dx}\:\:{and}\:\:\frac{\int{f}\left({x}\right){dx}}{\int{g}\left({x}\right){dx}}\:. \\ $$ Commented by maxmathsup by imad last updated on 16/Nov/18…

Question-178922

Question Number 178922 by mnjuly1970 last updated on 22/Oct/22 Answered by ARUNG_Brandon_MBU last updated on 22/Oct/22 $$\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{n}+\mathrm{2}} \\ $$$$\left(\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{2}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{4}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} }+\mathrm{2}\left(\frac{−\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)}+\frac{−\mathrm{2}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}+\frac{\mathrm{1}}{{n}\left({n}+\mathrm{2}\right)}\right)\right)…