Question

1 23 sin(43) dy dx by reversing the order of Evaluate Jo JC2 integration (1 –cos(1) ]](1+cos(1) (1 – cos(1) *] (1 – cos(1) (1+cos(1)

Answer 1

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dhe s теплом e sin () dy dx и y = x x=0 to y=1 to x=1. y! JA CI, X-o 녀 (0,0)

Now for coordinates of point A, put y=1 in y = x - x - 21 Here x20 so x = 1 So A= (1,1)

Now drawing horizontal stup. ay=1 Here y=x sezo - a- & Sy but in first quadrant, se so so x= sy So limits are x=0 to x= dy y=0 to y=1 Now dos = 2 sin (y) dyda & sinly) dedy Y=0 x= =0

ople-ch, je -s aint) ay - 1 s & sin (18) dy Now put yet & 3ydy=dt e & dy dit when yo, t= y² = (0)2 = 0 when y=1, t= y3 = 118 = 1 no s'yº sin (9) dy yo - I Ý sint) dt t=0

1 (cash) - (- ces (0) - Frosa) + cos(0) to [-c0111] because cos (0) = 1 so Solda & sin(y) dy dx - [-conto First option