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Question

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

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Solution

21 (e3x+7x5)dx

Here, a=1, b=2, nh=2(1)=3

=limh0 h[f(a)+f(a+h)+f(a+2h)+...+f(a+(¯¯¯¯¯¯¯¯¯¯¯¯¯n1))h]

f(a)=f(1)=e375

f(a+h)=f(h1)=e3(h1)+7(h1)5

f(a+2h)=f(2h1)=e3(2h1)+7(2h1)5

f(a+¯¯¯¯¯¯¯¯¯¯¯¯¯n1h)=f(¯¯¯¯¯¯¯¯¯¯¯¯¯n1h1)

=e3(¯¯¯¯¯¯¯¯¯n1h1)+7(¯¯¯¯¯¯¯¯¯¯¯¯¯n1h1)5

=limh0 h[(e375)+(e3h.e3+7h75)+(e6h.e3+14h75)+...+e3(n1)h.e3+7(n1) h75]

=limh0 h[e3 (1+e3h+e2(3h)+...+e3(n1)h)7n5n+7h+2(7h)+...+ (n1) 7h]

=limh0 h[e3((e3h)n1e3h1)7n5n+7h((n1)×n)2]

=limh0 [e3.e3nh13×e3h13h7nh5nh+72(nhh)(nh)]

=[13e3×e9117(3)5(3)+72(30)(3)]

=e913e32115+72×9

=e913e3+63236

=e913e3+63722

=e913e392

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