User: smitapn612

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Posts by smitapn612

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... $\text{Let I(a) ] } = \int_0^{\infty} \frac{ tan^{-1}ax }{x(1+x^2)} dx$ $\frac{dI}{da} = \int_0^{\infty} \frac{\partial}{\partial a} \frac{ tan^{-1}ax }{x(1+x^2)} dx$ $= \int_0^{\infty} \frac{ 1}{x(1+x^2)} \frac{x }{(1+x^2)} dx$ $= \frac{1}{1-a^2} \int_0^{\infty} [ \frac{ 1}{(1+x^2)} - ... written 13 months ago by smitapn6120 • updated 12 months ago by awari.swati831150 1 answer 336 views 1 answers ...$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 13 months ago by smitapn6120 • updated 12 months ago by awari.swati831150
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... $\text{ The region of integration is from y = 0 to y = } \sqrt{2x-x^2} \Longrightarrow x^2 + y^2 -2x = 0 \\$ $\text{x= 0 to x = 2} \\$ $\text{changing to polar coordinates x = } rcos\theta , y= rsin\theta, dxdy = rdrd\theta \\$ $I = \int_{\theta=0}^{\frac{\pi}{2}} \int_{r=0}^{2cos\thet ... written 13 months ago by smitapn6120 • updated 12 months ago by awari.swati831150 1 answer 273 views 1 answers ...$ \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}{2}} ...
written 13 months ago by smitapn6120 • updated 12 months ago by awari.swati831150
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... $\text { The limits of y are }y= x^2 \text{ and } y = 2-x \Longrightarrow x+y =2 \text { and x=0 to x=1}$ $\text { The point of intersection of x+y = 2 and y = } x^2$ $x + x^2 = 2 \Longrightarrow x^2 +x-2 = 0 \Longrightarrow \text{x=1, x=-2}$ The region of integration is OABC. To chan ...
written 13 months ago by smitapn6120 • updated 12 months ago by awari.swati831150
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... $\text { The limits for y are 0 and 1 and for x are 0 and x = } \sqrt{1-y^2}) \Longrightarrow x^2 = 1-y^2 \Longrightarrow x^2 + y^2 = 1 \\$ $\text{ Hence, the region of integration is the first quadrant of the circle} x^2 + y^2 =1. \text{If we change the order of integration,y varies from 0 t ... written 13 months ago by smitapn6120 • updated 12 months ago by awari.swati831150 1 answer 359 views 1 answers ...$ \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 13 months ago by smitapn6120 • updated 12 months ago by awari.swati831150
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... $\text{ The point of intersection r =} 2sin\theta \text{ and r = } 2cos\theta \\$ $tan\theta = 1 \Longrightarrow \theta = \frac{\pi}{4} \\$ $Area_1 = \int_{\theta = 0}^\frac{\pi}{4} \int_{r= 0}^{ 2sin\theta} r dr d\theta\\$ $= \int_{\theta = 0}^\frac{\pi}{4} \left[\frac{r^2}{2} \right] ... written 13 months ago by smitapn6120 • updated 12 months ago by awari.swati831150 1 answer 409 views 1 answers ...$ \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\\ ...