y=1−2x+ex∫0xe−tf(t)dtdxdy=−2+e−x⋅exf(x)+ex∫0xe−tf(t)dtdxdy=−2+y+y+2x−1dxdy−2y=(2x−3)ye−2x=∫(2x−3)dx⋅e−2xye−2x=2−(2x−3)e−2x+∫e−2xdxye−2x=2−(2x−3)e−2x−21e−2x+cf(0)=1⇒c=1−23+21=0
y=−2(2x−3)−21y=−x+1x+y=1 area =21(1)(1)=21
JEE Main 2025 — Mathematics Calculus
Let f:[0,∞)→R be differentiable function such that f(x)=1−2x+∫0xex−tf(t)dt for all x∈[0,∞).
Then the area of the region bounded by y=f(x) and the coordinate axes is
Held on 4 Apr 2025 · Verified 6 Jul 2026.
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