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User: smitapn612

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smitapn6120
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Posts by smitapn612

<prev • 137 results • page 1 of 14 • next >
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Answer: A: Prove that $\int_{0}^{\infty}\frac{tan^{-1}ax}{x(1+x^2)}dx\,=\frac{\pi}{2}log(1+
... $ \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{ 1}{(1+x^2)} - ...
written 3 months ago by smitapn6120 • updated 10 weeks ago by awari.swati831160
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Answer: A: Prove that $\int_{0}^{\infty}\frac{e^{-x}-e^{-ax}}{x\,secx}dx\,=\frac{1}{2}log\l
... $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 $ $ = \left [ \frac{ e^{-ax} }{a^ ...
written 3 months ago by smitapn6120 • updated 11 weeks ago by awari.swati831160
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Answer: A: Change to polar co-ordinates and evaluate $\int_{0}^{2}\int_{0}^{\sqrt{2x-x^2}}\
... $ \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}^{2cos\thet ...
written 3 months ago by smitapn6120 • updated 10 weeks ago by awari.swati831160
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Answer: A: By transforming to polar co-ordinates $\iint \frac{(x^2+y^2)}{(x^2+y^2)^{3/2}}dx
... $ \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}{2}} ...
written 3 months ago by smitapn6120 • updated 10 weeks ago by awari.swati831160
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Answer: A: Evaluate $\int_{0}^{1}\int_{x^2}^{2-x}xy\,dy\,dx$ by changing the order of integ
... $ \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} $ The region of integration is OABC. To chan ...
written 3 months ago by smitapn6120 • updated 10 weeks ago by awari.swati831160
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Answer: A: Evaluate $\int_{0}^{1}\int_{0}^{\sqrt{1-y^2}}\frac{cos^{-1}x}{\sqrt{1-x^2}\,\sqr
... $ \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 from 0 t ...
written 3 months ago by smitapn6120 • updated 10 weeks ago by awari.swati831160
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Answer: A: Change the order of the integration in the following $\int_{0}^{1/2}\int_{0}^{\s
... $ \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{1}{2} ...
written 3 months ago by smitapn6120 • updated 10 weeks ago by awari.swati831160
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Answer: A: Evaluate $\iint r^3\, drd\theta$ over the area included between the circles $r=
... $ \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} \right] ...
written 3 months ago by smitapn6120 • updated 10 weeks ago by awari.swati831160
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Answer: A: Evaluate $\iint_R y\,dxdy$ where 'R' is the region bounded by $y^2=4x$ & $x^2=4y
... $ \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= 4\\ ...
written 3 months ago by smitapn6120 • updated 10 weeks ago by awari.swati831160
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Answer: A: Evaluate $\iint_R\frac{2xy^5}{\sqrt{1+x^2y^2-y^4}}dxdy$ where 'R' is the region
... $ \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^5x} {\ ...
written 3 months ago by smitapn6120 • updated 10 weeks ago by awari.swati831160

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