
A1=∫0π/4(cosx−sinx)dx
A1=(sinx+cosx)0π/4=2−1
A2=∫0π/4sinxdx+∫π/4π/2cosxdx
=(−cosx)0π/4+(sinx)π/4π/2
A2=(1−21)+(1−21)=2−2
A1:A2=1:2,A1+A2=1
JEE Main 2021 — Mathematics Calculus
Let A1 be the area of the region bounded by the curves y=sinx,y=cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y=sinx,y=cosx, x-axis and x=2π in the first quadrant. Then,
Held on 26 Feb 2021 · Verified 6 Jul 2026.
2A1=A2 and A1+A2=1+2
A1:A2=1:2 and A1+A2=1
A1:A2=1:2 and A1+A2=1
A1=A2 and A1+A2=2
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