find-the-function-f-x-satisfying-the-given-conditions-i-f-x-4x-2-1-f-0-3-ii-f-x-12-f-0-2-f-0-3-iii-f-x-2x-f-0-3-f-0-2-help-me-sir-pleas-

Question Number 94245 by mhmd last updated on 17/May/20

$$findthefunctionf\left(x\right)satisfyingthegivenconditions$$$$\left(i\right)f^{{}^{\prime }}\left(x\right)=4x^2-1,f\left(0\right)=3?$$$$\left(ii\right)f^{{}^{″}}\left(x\right)=12,f^{{}^{\prime }}\left(0\right)=2,f\left(0\right)=3?$$$$\left(iii\right)f^{{}^{″}}\left(x\right)=2x,f^{{}^{\prime }}\left(0\right)=-3,f\left(0\right)=2?$$$$$$$$helpmesirpleas?$$

Answered by Ar Brandon last updated on 17/May/20

$$\mathrm{Q}1\mathrm{\setminus }{\int }_0^1{\int }_0^1\frac{1}{\sqrt{1+{\mathrm{x}}^2-{\mathrm{y}}^2}}\mathrm{dydx}={\int }_0^1\frac{1}{\sqrt{1+{\mathrm{x}}^2}}{\int }_0^1\frac{1}{\sqrt{1-{\left(\frac{\mathrm{y}}{\sqrt{1+{\mathrm{x}}^2}}\right)}^2}}\mathrm{dydx}$$$$={\int }_0^1{\left[{\sin }^{-1}\left(\frac{\mathrm{y}}{\sqrt{1+{\mathrm{x}}^2}}\right)\right]}_0^1\mathrm{dx}={\int }_0^1{\sin }^{-1}\left(\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\right)\mathrm{dx}$$$$\mathrm{u}={\sin }^{-1}\left(\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\right)\Rightarrow \mathrm{du}=-2\mathrm{x}\cdot \frac{1}{2}\cdot \frac{1}{{(1+{\mathrm{x}}^2)}^{\frac{3}{2}}}\cdot \frac{1}{\sqrt{1-\frac{1}{1+{\mathrm{x}}^2}}}$$$$=-\frac{\mathrm{x}}{{(1+{\mathrm{x}}^2)}^{\frac{3}{2}}}\cdot \frac{1}{\sqrt{\frac{{\mathrm{x}}^2}{1+{\mathrm{x}}^2}}}=-\frac{\mathrm{x}}{{(1+{\mathrm{x}}^2)}^{\frac{3}{2}}}\cdot \frac{\sqrt{1+{\mathrm{x}}^2}}{\mathrm{x}}=-\frac{1}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{dv}=\mathrm{dx}\Rightarrow \mathrm{v}=\mathrm{x}$$$${\int }_0^1{\sin }^{-1}\left(\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\right)\mathrm{dx}={\left[{\mathrm{xsin}}^{-1}\left(\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\right)\right]}_0^1+{\int }_0^1\frac{\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=\frac{\pi }{4}+{\left[\frac{1}{2}\ln (1+{\mathrm{x}}^2)\right]}_0^1=\frac{\pi }{4}+\frac{1}{2}\ln \left(2\right)$$$$$$

Commented by mhmd last updated on 17/May/20

$$thankyousir$$

Commented by Ar Brandon last updated on 17/May/20

��You're welcome. I realised the previous post was deleted and that's why I decided to post it here.

Commented by mhmd last updated on 17/May/20

$$idontknowsir?imdontdeleted$$

Commented by Tinku Tara last updated on 17/May/20

$$\mathrm{Did}\mathrm{the}\mathrm{post}\mathrm{got}\mathrm{deleted}\mathrm{after}\mathrm{an}$$$$\mathrm{answer}\mathrm{or}\mathrm{comment}\mathrm{added}?$$$$\mathrm{That}\mathrm{is}\mathrm{disabled}\mathrm{from}\mathrm{server}.$$

Commented by mhmd last updated on 17/May/20

$$sircanhelpmeknthequestion94243?$$

Commented by Ar Brandon last updated on 17/May/20

I think it was deleted before I commented. Because after solving I tried to send but to no avail so I gave up. But my solution was saved. And when I wanted to answer the next question it just appeared so I sent it.

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