JEE Mains · Maths · STD 12 - 7.2 definite integral
\(\lim _{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}\) is equal to
- A \(12\)
- B \(\frac{19}{3}\)
- C \(0\)
- D \(19\)
Answer & Solution
Correct Answer
(D) \(19\)
Step-by-step Solution
Detailed explanation
\(\lim _{n \rightarrow \infty} \frac{3}{n} \sum \limits_{r=0}^{n-1}\left(2+\frac{r}{n}\right)^2\) \(=3 \int \limits_0^1(2+x)^2 d x=27-8=19\)
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