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Question

# Evaluate: ∫π20sin2xsinx+cosxdx. OR Evaluate: ∫2−1(e3x+7x−5)dx as a limit of sums.

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Solution

## Let I=∫π20sin2xsin x+cos xdx......(i) ⇒I=∫π20sin2(π2−x)sin(π2−x)+cos(π2−x)dx.. [Using Property, ∫a0f(x)dx=∫a0f(a−x)dx] ⇒I=∫π20cos2xsin x+cos xdx.......(ii)Adding, (i) and (ii),⇒2I=∫π20sin2x+cos2xsin x+cos xdx⇒2I=∫π20dxsin x+cos x⇒2I=1√2∫π20dxsin x.1√2+cos x1√2⇒2I=1√2∫π20dxsin x.cosπ4+cos x.sinπ4 ⇒2I=1√2∫π/20cosec(π4+x)dx⇒2I=1√2[ln|cosec(π4+x)−cotcosec(π4+x)|]π/20⇒2I=1√2[ln|cosec(π4+π2)−cot(π4+π2)|]−[ln|cosec(π4+0|)−cot(π4+0)]⇒2I=1√2[ln|√2−(−1)|−ln|√2−1]⇒=12√2[ln|√2+1√2−1|] OR ∫2−1(e3x+7x−5)dx.Here f(x)=e3x+7x−5a=−1,b=2,h=b−an=3nBy definition∫2−1(e3x+7x−5)dx=limn→∞∑nr=1h.f(a+rh)limn→∞∑nr=1h.f(−1+rh)=limn→∞∑nr=1h.(e3(−1+rh)+7(−1+rh)−5) limn→∞[h.e−3.e3h(1+e3h+e6h+.....+e3nh)+7h2(1+2+3+.....+n)−12nh] limn→∞[he3he3×e3nh−1e3h−1+7h2n(n+1)2−12nh] limn→∞[(he3×3nne3)×(e3×3n−1)×n3×3+63n2×n(n+1)2−12×3] Now applying the limit we get =e9−13e3+632−36=e9−13e3−92

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