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-- Mathematics - Section B1
Definite Integration
If f(x) is differentiable function and f(x)=x2+x∫0e−t(x−t)dt, then f(−1) equals to |
−23 23 13 −13 |
23 |
We have, f(x)=x2+x∫0e−t(x−t)dt ......(i) ⇒f(x)=x2+x∫0e−(x−t)f(x−(x−t))dt ⇒f(x)=x2+x∫0e−xetf(t)dt ⇒f(x)=x2+e−xx∫0etf(t)dt ⇒f′(x)=2x+e−x{exf(x)}−e−xx∫0etf(t)dt ⇒f′(x)=2x+f(x)−x∫0e−(x−t)f(t)dt ⇒f′(x)=2x+f(x)−x∫0e−(x−(x−t))f(x−t)dt ⇒f′(x)=2x+f(x)−x∫0e−tf(x−t)dt ⇒f′(x)=2x+f(x)+x2−f(x) [Using (i)] ⇒f′(x)=2x+x2 ⇒f(x)=x2+x33+C ....(ii) From (i), we obtain f(0)=0. Puting x=0, in (ii), we obtain C=0 ∴ f(x)=x2+x33 ⇒f(−1)=1−13=23 |