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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > if u r genius solve this problem y=1/(x^2); find...
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if u r genius solve this problem y=1/(x^2); find the area between -1 to 1. by Abdul.

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Iitjee/olympiad/kvpy/foundation/neet/boards/fe/se crash course and regular course. Experience 17+ yrs

Function is not defined at x =0
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Teaching maths from 7 years

Integrate both side with respect to x then take the limit from -1 to +1, then put the limit after that you will get area=0.
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senior professor and trainer in maths

doesn't exist
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"Decoding the World of Physics and Math: 12 Years of Expertise, Powered by a Teaching Enthusiasts"

You integrate from 0 to 1 and than multiply that area with 2 or integrate from -1 to 0- and take its mode because area is positive quantity and than integrate function from 0+ to 1 and add both areas.
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Mathematics for JEE Mains/Advanced, XI & XII (All Boards)

The integral is non-convergent that is it will tend to infinity. Why? Just Draw the graph of 1/x2. You will understand. Also on solving by integration you have to break the intervals from -1 to 0 and 0 to 1 as the curve is discontinuous. Now while taking the limit of 1/x with x -> inf you will find ans...
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The integral is non-convergent that is it will tend to infinity. Why? Just Draw the graph of 1/x2. You will understand. Also on solving by integration you have to break the intervals from -1 to 0 and 0 to 1 as the curve is discontinuous. Now while taking the limit of 1/x with x -> inf you will find ans tends to inf. read less
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The x and y axes are both asemptodes to the curve.. The curve touches the y axis at infinity. So area is infinity.
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integrate 1/(x^2) between limits -1 to +1 , integrating 1/x^2 will be -(1/X). Integrating between -1 to +1 will be ( -1/1-(-1/-1)) = (-1-1)=-2
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IIT JEE Trainer

in fact the given function is not bounded in the limit -1 to 1,so if you see the graph of the function it seem that y axis is the asymptotic to the given curve, for more detail refer any book with title mathematics for dummy's or mathematics demystified by any author. chirag
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