Correct answer: (C) limn→∞1/n6n+1^2+2^2+…+9n^2n^2. Let 2+x^2=f(x) ⇒∫032+x^2dx=∫03fxdx Now, ∫03fxdx=limn→∞1/n∑r=13nfr/n =limn→∞1/n∑r=13n2+rn^2 =limn→∞1/n2×3n+1^2+2^2+3^2+…+3n^2n^2 =limn→∞1/n6n+1^2+2^2+3^2+…+3n^2n^2

Q116 / 160 · TS EAMCET 2018 (04 May Shift 1) · Mathematics

$\int_{0}^{3} (2 + x^{2}) d x =$

A
$\lim_{n \to \infty} \frac{1}{n} [2 n + \frac{1^{2} + 2^{2} + \dots + {(3 n)}^{2}}{n^{2}}]$
B
$\lim_{n \to \infty} \frac{1}{n} [3 n + \frac{1^{2} + 2^{2} + \dots + 6 n^{2}}{n^{2}}]$
Correct answer
C
$\lim_{n \to \infty} \frac{1}{n} [6 n + \frac{1^{2} + 2^{2} + \dots + 9 n^{2}}{n^{2}}]$
D
$\lim_{n \to \infty} \frac{1}{n} [3 n + \frac{1^{2} + 2^{2} + \dots + 3 n^{2}}{n^{2}}]$

Solution

Let $2 + x^{2} = f (x)$

$\Rightarrow \int_{0}^{3} (2 + x^{2}) d x = \int_{0}^{3} f (x) d x$

Now, $\int_{0}^{3} f (x) d x = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{3 n} f (\frac{r}{n})$

$= \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{3 n} (2 + {(\frac{r}{n})}^{2})$

$= \lim_{n \to \infty} \frac{1}{n} (2 \times 3 n + \frac{1^{2} + 2^{2} + 3^{2} + \dots + {(3 n)}^{2}}{n^{2}})$

$= \lim_{n \to \infty} \frac{1}{n} (6 n + \frac{1^{2} + 2^{2} + 3^{2} + \dots + {(3 n)}^{2}}{n^{2}})$

Hence, the correct answer is option: 'C'.

∫032+x2dx=

Solution

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$\int_{0}^{3} (2 + x^{2}) d x =$