IIT JAM MATHEMATICS 2019 | Real Analysis | Double Integral | Q 9

Welcome friends, in this video we are going to discuss the problem of IIT J&M 2019 Problem of Real Analysis, Question number 9 the value of the integral y equal to 0 to y equal to 1 x is equal to 0 to 1 minus y square y into sin pi 1 minus x power 2 dx dy we have four option first option a equal to 1 upon 2 pi b equal to 2 pi c equal to pi by 2 d equal to 2 by pi that's the problem of evaluation of double integral so if you integrate sin pi 1 minus x square with respect to x so it's difficult term to integrate therefore we change the order of integration and for that we require the region of integration so region of integration is bounded by these four curve or four limits x equal to 0 x is equal to 1 minus y square y equal to 0 and y equal to 1 so x is equal to 0 is a y-axis x is equal to 1 minus y square so this is a curve x is equal to 1 minus y square y equal to 0 x-axis and y equal to 1 so region which is bounded by these four curves is given by this line and we want to change the order of integration from dx dy to dy dx for that we have to draw one vertical arrow in the region so this initial point will give the limit of y lower limit of y and terminal point will give the upper limit of y so lower limit is y equal to 0 and upper limit is this curve x is equal to 1 minus y square if we buy the subject of the equation then y equal to square root 1 minus x and the limit for x is minimum value of x in the region which is x equal to 0 and maximum value of x in the region which is x is equal to 1 so limits for x is x equal to 0 to x is equal to now we are going to evaluate this integration so we integrate with respect to y so y square by 2 evaluate between the limit 0 to square root 1 minus x after substituting the lower limits and upper limits we have 1 minus x and 1 by 2 is constant so take out from the integration now to evaluate this integration we'll use the substitution pi 1 minus x square equal to t so differentiate both side with respect to x then 2 pi 1 minus x minus dx equal to dt so 1 minus x dx is equal to minus dt by 2 pi now limits for x is equal to 0 to 1 if you put x is equal to 0 then t equal to pi and if you put x is equal to 1 then t equal to 0 so limits for t are pi to 0 now substitute all this value in the integration so 1 by 2 limit t equal to pi to t equal to 0 sin t and 1 minus x dx is equal to minus dt by 2 pi now integrates sin t which is minus cos t evaluated between the limit 0 to pi then we have minus cos pi plus cos 0 so minus cos pi is value minus 1 minus into minus plus 1 1 plus 1 equal to 2 2 divided by 4 pi is equal to 1 upon 2 pi and which is the option number one it is the correct answer thanks for watching for more videos like share subscribe press the bell icon of next gen ms