23Ï€ab if i remember ..
see find area in first quadrant , i.e ∫ydx , then by symmetry total area = 4 times of that
find the area of the ellipse given by the equation x2a2 + y2b2 = 1
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3 Answers
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21thnx ur ri8..i also got the same answer but the doubt is
we know f(x) = √(1-x2 gives a circle of area pi
let -1 to +1 ∫f(x)dx = pi
we know \int_{ca}^{cb}{}f(t)dt = c\int_{a}^{b}{}f(ct)dt
b{f(x/a)} gives the ellipse
using above property b\int_{-a}^{a}{}f(x/a) = ab\int_{-1}^{1}{}f(x)
which gives area of ellipse equal to pi ab
but -1 to +1 ∫f(x)dx ≠pi
but still we are getting the area correct,why?
23if u r taking f = √1-x2 , then that is the curve only above the x axis , so you are calculating only half of the area when u r writing -1 to +1 ∫f(x)dx
same for ellipse, u r calculating only half of ellipse's area