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Evaluate the following as the limit of sums

∫2(x2+3) dx

0 2 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Short answer type Questions as classified in NCERT Exemplar

    Sol:

    Let    I=∫02(x2+3) dx  Using the  formula,∫abf(x) dx=limh→0h[f(a)+f(a+h)+f(a+2h)+…+f(a+n−1¯h)]where  h=b−an            Here,  a=0  and  b=2∴          h=2−0n            ∴nh=2Here,     f(x)=x2+3                 f(0)=02+3=0+3=3          f(0+h)=(0+h)2+3=h2+3        f(0+2h)=(0+2h)2+3=4h2+3=3………………………………f(0+n−1¯h)=(0+n−1¯h)2+3(n−1)2h2+3Now,  ∫02(x2+3) dx=limh→0h[3+h2+3+4h2+3+…+(n−1)2h2+3]                                        =limh→0h[(3+3+3+…+n)+{h2+4h2+…+(n−1)2h2}]                                        =limh→0h[3n+h2{1+4+…+(n−1)2}]                                        =limh→0h[3n+h2n(n−1)(2n−1)6]                                                                         [?1+4+9+…+(n−1)2=n(n−1)(2n−1)6]                                        =limh→0[3nh+h3n(n−1)(2n−1)6]                                        =limh→0[3nh+nh(nh−h)(2nh−h)6]                                        =[3×2+2(2−0)(2×2−0)6]          [?nh=2    and    h=0]                                        =[6+2×2×46]=6+83=263Hence,  ∫02(x2+3) dx=26

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