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Author: Tinku Tara

Question-205315

Question Number 205315 by cherokeesay last updated on 15/Mar/24 Answered by MM42 last updated on 16/Mar/24 $$\left[{x}\right]+\left[{x}−\frac{\mathrm{1}}{\mathrm{2}}\right]=\left[\mathrm{2}{x}\right]−\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\left[{x}\right]} \left(\left[\mathrm{2}{x}\right]−\mathrm{1}\right){dx}=\left(\left[\mathrm{2}{x}\right]−\mathrm{1}\right){x}\mid_{\mathrm{0}} ^{\left[{x}\right]} \\ $$$$=\left(\left[\mathrm{2}{x}\right]−\mathrm{1}\right)\left[{x}\right] \\…

pi-2-pi-2-8-2-cosx-1-e-sinx-1-sin-4-x-dx-api-blog-3-2-2-then-find-a-b-

Question Number 205279 by gopikrishnan last updated on 14/Mar/24 $$\underset{-\pi/\mathrm{2}} {\int}^{\pi/\mathrm{2}} \frac{\mathrm{8}\sqrt{\mathrm{2}}{cosx}}{\left(\mathrm{1}+\overset{{sinx}} {{e}}\right)\left(\mathrm{1}+{si}\overset{\mathrm{4}} {{n}x}\right)}{dx}={a}\pi+{blog}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:{then}\:{find}\:{a}+{b} \\ $$ Answered by Berbere last updated on 14/Mar/24 $${x}\rightarrow−{x} \\…

If-f-x-1-2x-2-Find-f-x-

Question Number 205273 by hardmath last updated on 14/Mar/24 $$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{2x}+\mathrm{2} \\ $$$$\mathrm{Find}\:\:\:\mathrm{f}\left(\mathrm{x}\right)=? \\ $$ Answered by Rasheed.Sindhi last updated on 14/Mar/24 $${Replace}\:\mathrm{x}\:{by}\:\mathrm{x}+\mathrm{1} \\ $$$$\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{2}=\mathrm{2x}+\mathrm{4} \\…

Question-205269

Question Number 205269 by cherokeesay last updated on 14/Mar/24 Answered by som(math1967) last updated on 14/Mar/24 $$\:\int_{−\mathrm{2}} ^{\mathrm{2}} \mathrm{2}{f}\left({x}\right){dx}\:\:\left[\:\because{f}\left({x}\right)={f}\left(−{x}\right)\right] \\ $$$$=\mathrm{2}\int_{−\mathrm{2}} ^{\mathrm{2}} {f}\left({x}\right){dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}}…

Question-205280

Question Number 205280 by cherokeesay last updated on 14/Mar/24 Commented by mr W last updated on 14/Mar/24 $${figure}\:{is}\:{not}\:{uniquely}\:{defined}. \\ $$$${or}\:{you}\:{mean}\:{that}\:{the}\:{hypotenuse}\:{of} \\ $$$${both}\:{triangles}\:{is}\:{of}\:{same}\:{length}? \\ $$ Commented…