Let f:[0,∞)→R be a differentiable function such that
f(x)=1−2x+∫x0ex−tf(t)dt for all x∈[0,∞).
Then the area of the region bounded by y=f(x) and the coordinate axes is
Solutions · 1
Official solution
∵f(x)=1−2x+∫x0ex−tf(t)dt
or, f(x)=1−2x+ex∫x0e−tf(t)dt
on differentiating both sides w.r.t. x we get
f′(x)=−2+ex∫x0e−tf(t)dt+ex⋅e−xf(x)f′(x)=−2+f(x)+2x−1+f(x)∴f(x)−2f(x)=2x−3=e∫−2dx=e−2x∴e−2x⋅f(x)=∫e−2x(2x−3)dxe−2x⋅f(x)=(2x−3)⋅−2e−2x−∫2⋅−2e−2xdxe−2x⋅f(x)=−2(2x−3)e−2x+−2e−2x+cf(x)=−x+1+c′e2x∵f(x)=1∴1=0+1+c′⇒c′=0∴f(x)=−x+1⇒ Area =21