Question
Mathematics Question on Binomial theorem
Find ∫02(x2+1)dx as the limit of a sum :
A
(A) 43
B
(B) 143
C
(C) 145
D
(D) None of these
Answer
(C) 145
Explanation
Solution
Explanation:
We know that:∫abf(x)dx=(b−a)limn→∞1n(f(a)+f(a+h)+…+f(a+(n−1)h))Putting a=0, b=2, h=b−an=2−0n=2nin ∫02x2+1dxI=(2−0)limn→∞1n(f(0)+f(n)+f(2n)+…+fn−1)hf(0)=1f(h)=h2+1=(4n2)+1f((n−1)h)=(n−1)2×4n2+1∴I=2limn→∞1n((1+1+…n times )+(0+4n2+16n2+…+(n−1)2n2))=2limn→∞1n(n+4n(n−1)n(2n−1)6)=2limn→∞(1+23(1−1n)(2−1n))=2×(1+43)=143Hence, the correct option is (C).