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 ?


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

$f i n d t h e f u n c t i o n f (x) s a t i s f y i n g t h e g i v e n c o n d i t i o n s$ $(i) f^{^{'}} (x) = 4 x^{2} - 1, f (0) = 3 ?$ $(i i) f^{^{″}} (x) = 12, f^{^{'}} (0) = 2, f (0) = 3 ?$ $(i i i) f^{^{″}} (x) = 2 x, f^{^{'}} (0) = - 3, f (0) = 2 ?$ $h e l p m e s i r p l e a s ?$
	Answered by Ar Brandon last updated on 17/May/20
	$Q1\ ∫_0 ^1 ∫_0 ^1 (1/(√(1+x^2 −y^2 )))dydx=∫_0 ^1 (1/(√(1+x^2 )))∫_0 ^1 (1/(√(1−((y/(√(1+x^2 ))))^2 )))dydx =∫_0 ^1 [sin^(−1) ((y/(√(1+x^2 ))))]_0 ^1 dx=∫_0 ^1 sin^(−1) ((1/(√(1+x^2 ))))dx u=sin^(−1) ((1/(√(1+x^2 ))))⇒du=−2x∙(1/2)∙(1/((1+x^2 )^(3/2) ))∙(1/( (√(1−(1/(1+x^2 )))))) =−(x/((1+x^2 )^(3/2) ))∙(1/( (√(x^2 /(1+x^2 )))))=−(x/((1+x^2 )^(3/2) ))∙((√(1+x^2 ))/x)=−(1/(1+x^2 ))dx dv=dx⇒v=x ∫_0 ^1 sin^(−1) ((1/(√(1+x^2 ))))dx=[xsin^(−1) ((1/(√(1+x^2 ))))]_0 ^1 +∫_0 ^1 (x/(1+x^2 ))dx =(π/4)+[(1/2)ln(1+x^2 )]_0 ^1 =(π/4)+(1/2)ln(2)$
	$Q 1 ∖ \int_{0}^{1} \int_{0}^{1} \frac{1}{\sqrt{1 + x^{2} - y^{2}}} dydx = \int_{0}^{1} \frac{1}{\sqrt{1 + x^{2}}} \int_{0}^{1} \frac{1}{\sqrt{1 - {(\frac{y}{\sqrt{1 + x^{2}}})}^{2}}} dydx$ $= \int_{0}^{1} {[\sin^{- 1} (\frac{y}{\sqrt{1 + x^{2}}})]}_{0}^{1} dx = \int_{0}^{1} \sin^{- 1} (\frac{1}{\sqrt{1 + x^{2}}}) dx$ $u = \sin^{- 1} (\frac{1}{\sqrt{1 + x^{2}}}) \Rightarrow du = - 2 x \cdot \frac{1}{2} \cdot \frac{1}{{(1 + x^{2})}^{\frac{3}{2}}} \cdot \frac{1}{\sqrt{1 - \frac{1}{1 + x^{2}}}}$ $= - \frac{x}{{(1 + x^{2})}^{\frac{3}{2}}} \cdot \frac{1}{\sqrt{\frac{x^{2}}{1 + x^{2}}}} = - \frac{x}{{(1 + x^{2})}^{\frac{3}{2}}} \cdot \frac{\sqrt{1 + x^{2}}}{x} = - \frac{1}{1 + x^{2}} dx$ $dv = dx \Rightarrow v = x$ $\int_{0}^{1} \sin^{- 1} (\frac{1}{\sqrt{1 + x^{2}}}) dx = {[{xsin}^{- 1} (\frac{1}{\sqrt{1 + x^{2}}})]}_{0}^{1} + \int_{0}^{1} \frac{x}{1 + x^{2}} dx$ $= \frac{π}{4} + {[\frac{1}{2} \ln (1 + x^{2})]}_{0}^{1} = \frac{π}{4} + \frac{1}{2} \ln (2)$
	Commented by mhmd last updated on 17/May/20

	$t h a n k y o u s i r$
	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

	$i d o n t k n o w s i r ? i m d o n t d e l e t e d$
	Commented by Tinku Tara last updated on 17/May/20

	$Did the post got deleted after an$ $answer or comment added ?$ $That is disabled from server .$
	Commented by mhmd last updated on 17/May/20

	$s i r c a n h e l p m e k n t h e q u e s t i o n 94243 ?$
	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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