Let f(x)=ex ∴∫01f(x)g(x)dx=∫01ex(x2−ex)dx =∫01x2exdx−∫01e2xdx =[x2ex]01−2[xex−ex]01−21[e2x] =e−[2e2−21]−2[e−e+1]=e−2e2−23
Let f(x) be a function satisfying f′(x)=f(x) with f(0)=1 and g(x) be a function that satisfies f(x)+g(x)=x2. Then the value of the integral ∫01f(x)g(x)dx, is
Held on 30 Apr 2003 · Verified 6 Jul 2026.
e+2e2+25
e−2e2−25
e+2e2−23
e−2e2−23
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