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Question
Evaluate, ∫21(3x2−1)dx as the limit of a sum
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Solution
Consider the problem
Given ∫21(3x2−1)dx
Consider x−1+rh
where,
h=2−1n=1n
As, x→1,r→0
And x→2,r→1n
∫21(3x2−1)dx=limn→∞∑nr=0h[3(1+rh)2−1]=limn→∞∑nr=11n[3(1+rn)2−1]=limn→∞∑nr01n((3+3r2n2+6rn)−1)=limn→∞∑nr1n(2+3r2n2+6rn)=limn→∞1n(∑nr=0(2)+∑nr=0(3r2n2)+∑nr=0(6r2))=limn→∞1n(∑nr=0(2)+3n2∑nr=0(r2)+6n∑nr=0(r))=limn→∞1n(2n+3n2(n(n+1)(2n+1)6)+6n(n(n+1)2))=limn→∞(2nn+3n3(n(n+1)(2n+1)6)+6n2(n(n+1)2))=limn→∞(2+1n3(n(n+1)(2n+1)2)+3n2(n(n+1)1))=limn→∞(2+12(1)(1+1n)(2+1n)+3(1)(1+1n))=2+12limn→∞((1+1n)(2+1n)+3limn→∞(1+1n))=2+12×2+3×1=2+1+3=6
Hence, ∫21(3x2−1)dx=6
Given ∫21(3x2−1)dx
Consider x−1+rh
where,
h=2−1n=1n
As, x→1,r→0
And x→2,r→1n
∫21(3x2−1)dx=limn→∞∑nr=0h[3(1+rh)2−1]
=limn→∞∑nr=11n[3(1+rn)2−1]=limn→∞∑nr01n((3+3r2n2+6rn)−1)=limn→∞∑nr1n(2+3r2n2+6rn)=limn→∞1n(∑nr=0(2)+∑nr=0(3r2n2)+∑nr=0(6r2))=limn→∞1n(∑nr=0(2)+3n2∑nr=0(r2)+6n∑nr=0(r))=limn→∞1n(2n+3n2(n(n+1)(2n+1)6)+6n(n(n+1)2))=limn→∞(2nn+3n3(n(n+1)(2n+1)6)+6n2(n(n+1)2))=limn→∞(2+1n3(n(n+1)(2n+1)2)+3n2(n(n+1)1))=limn→∞(2+12(1)(1+1n)(2+1n)+3(1)(1+1n))=2+12limn→∞((1+1n)(2+1n)+3limn→∞(1+1n))=2+12×2+3×1=2+1+3=6
Hence, ∫21(3x2−1)dx=6
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