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## User: smitapn612

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#### Posts by smitapn612

<prev • 137 results • page 1 of 14 • next >
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...  $\text{Let I(a) ] } = \int_0^{\infty} \frac{ tan^{-1}ax }{x(1+x^2)} dx \\$ $\frac{dI}{da} = \int_0^{\infty} \frac{\partial}{\partial a} \frac{ tan^{-1}ax }{x(1+x^2)} dx \\$ $= \int_0^{\infty} \frac{ 1}{x(1+x^2)} \frac{x }{(1+x^2)} dx \\$ $= \frac{1}{1-a^2} \int_0^{\infty} [ \frac{ ... written 21 days ago by smitapn6120 1 answer 55 views 1 answers ... $ \text{Let I(a) } = \int_0^{\infty} \frac{ (e^{-x} - e^{-ax}) }{xsecx} dx \\  \frac{dI}{da} = \int_0^{\infty} \frac{\partial}{\partial a} \frac{ (e^{-x} - e^{-ax}) }{xsecx} dx \\  = \int_0^{\infty} \frac{ x e^{-ax} }{xsecx} dx \\  = \int_0^{\infty} e^{-ax} cosx dx \\  = \ ...
written 21 days ago by smitapn6120 • updated 10 days ago by Saurabh Singh0
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...  $\text{ The region of integration is from y = 0 to y = } \sqrt{2x-x^2} \Longrightarrow x^2 + y^2 -2x = 0 \\$ $\text{x= 0 to x = 2} \\$ $\text{changing to polar coordinates x = } rcos\theta , y= rsin\theta, dxdy = rdrd\theta \\$ $I = \int_{\theta=0}^{\frac{\pi}{2}} \int_{r=0}^{2co ... written 21 days ago by smitapn6120 • updated 10 days ago by Saurabh Singh0 1 answer 48 views 1 answers ... $ \text{ The circle } x^2 + y^2 - 2ax = 0 \text { is a circle with center (a,0) and radius a. To change to polar coordinate.} \\  x= rcos\theta, y = rsin\theta \\  dxdy = rdrd\theta \\  \text {The equation of circle becomes r = acos}\theta \\  I = \int_{\theta=0}^{\frac{\pi ...
written 21 days ago by smitapn6120
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...  $\text { The limits of y are }y= x^2 \text{ and } y = 2-x \Longrightarrow x+y =2 \text { and x=0 to x=1} \\$ $\text { The point of intersection of x+y = 2 and y = } x^2 \\$ $x + x^2 = 2 \Longrightarrow x^2 +x-2 = 0 \Longrightarrow \text{x=1, x=-2} \\$ $\text { The region of inte ... written 21 days ago by smitapn6120 • updated 10 days ago by Saurabh Singh0 1 answer 55 views 1 answers ... $ \text { The limits for y are 0 and 1 and for x are 0 and x = } \sqrt{1-y^2}) \Longrightarrow x^2 = 1-y^2 \Longrightarrow x^2 + y^2 = 1 \\  \text{ Hence, the region of integration is the first quadrant of the circle} x^2 + y^2 =1. \text{If we change the order of integration,y varies fr ...
written 22 days ago by smitapn6120
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...  $\text { The region of integration is bounded by x = 0 i.e. the y-axis } \\$ $x = \sqrt{1-4y^2} \Longrightarrow x^2 = 1- 4y^2 \\$ $x^2 + \frac{y^2} {\frac{1}{4}} = 1 \\$ $\text { It is an ellipse with semi major axis 1 and semi minor axis } \frac{1}{2} \text {y = 0 and y = } \frac ... written 22 days ago by smitapn6120 • updated 10 days ago by Saurabh Singh0 1 answer 61 views 1 answers ... $ \text{ The point of intersection r =} 2sin\theta \text{ and r = } 2cos\theta \\  tan\theta = 1 \Longrightarrow \theta = \frac{\pi}{4} \\  Area_1 = \int_{\theta = 0}^\frac{\pi}{4} \int_{r= 0}^{ 2sin\theta} r dr d\theta\\  = \int_{\theta = 0}^\frac{\pi}{4} \left[\frac{r^2}{2} \ ...
written 22 days ago by smitapn6120
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...  $\text { The two parabola's intersect at O(0,0) and A (4,4).} \\$ $\text {Point of intersection : } \\$ $y^2 = 4x , x^2 = 4y \\$ $x= \frac{y^2}{4} \\$ $x^2 = 4y \\$ $\frac { y^4} {16} = 4y \\$ $y^4 = 64 y \\$ $y(y^3- 64) = 0 \\$ $y= 0 , y^3-64 = 0 \text { i.e. } y ... written 22 days ago by smitapn6120 1 answer 58 views 1 answers ... $ \text { Let O (0,0), A (1,1), B (0,1) be the vertices of the triangle OAB.} \\  \text { The equation of the line OA is} \\  \frac{x-0}{0-1} = \frac{y-0}{0-1} \text{ i.e. } x= y \\  \text {Now consider a strip parallel to x-axis } \\  \int_{y=0}^{1} \int_{x=0}^{y} \frac{2y^ ...
written 22 days ago by smitapn6120

#### Latest awards to smitapn612

Centurion 27 days ago, created 100 posts.
Rising Star 27 days ago, created 50 posts within first three months of joining.

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