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Question

Evaluate: π20sin2xsinx+cosxdx.

OR

Evaluate: 21(e3x+7x5)dx as a limit of sums.

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Solution

Let I=π20sin2xsin x+cos xdx......(i)

I=π20sin2(π2x)sin(π2x)+cos(π2x)dx.. [Using Property, a0f(x)dx=a0f(ax)dx]

I=π20cos2xsin x+cos xdx.......(ii)Adding, (i) and (ii),2I=π20sin2x+cos2xsin x+cos xdx2I=π20dxsin x+cos x2I=12π20dxsin x.12+cos x122I=12π20dxsin x.cosπ4+cos x.sinπ4

2I=12π/20cosec(π4+x)dx2I=12[ln|cosec(π4+x)cotcosec(π4+x)|]π/202I=12[ln|cosec(π4+π2)cot(π4+π2)|][ln|cosec(π4+0|)cot(π4+0)]2I=12[ln|2(1)|ln|21]=122[ln|2+121|]

OR

21(e3x+7x5)dx.Here f(x)=e3x+7x5a=1,b=2,h=ban=3nBy definition21(e3x+7x5)dx=limnnr=1h.f(a+rh)limnnr=1h.f(1+rh)=limnnr=1h.(e3(1+rh)+7(1+rh)5)

limn[h.e3.e3h(1+e3h+e6h+.....+e3nh)+7h2(1+2+3+.....+n)12nh]

limn[he3he3×e3nh1e3h1+7h2n(n+1)212nh]
limn[(he3×3nne3)×(e3×3n1)×n3×3+63n2×n(n+1)212×3]

Now applying the limit we get

=e913e3+63236=e913e392


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