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Indian_Dragon (95)

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Olaaa!! Perrrfect answer. 17  [22 rates]

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look if you put a value less than -1 for x the graph does not exist becoz y^2 becomes -ve so the graph exists b/w 0 and -1 in the second or third qd.
now the graph is symmetrical about the x axis. so what ever be the graph in 2nd and 3rd qd. . you can put limits from -1 to 0 for dx and integrate it and then double it.
that would be the required area.
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