If _n → ∞ _k=1^n ≤ft[ ≤ft( 3k/n )^2 + 2 ] 3/n = _0^6 f(x) dx , then: - Mathematics MCQ Mathematics Mock Test - 4 2026 | ExamTest.live

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If lim_{n \to \infty} \sum_{k = 1}^{n} [(\frac{3 k}{n})^{2} + 2] \frac{3}{n} = \int_{0}^{6} f (x) d x, then:

A $f (0) = 1$
B $f (x) = \frac{1}{3} x^{2} - \frac{1}{2} x + 1$
C $lim_{n \to \infty} \sum_{k = 1}^{n} [(\frac{3 k}{n})^{2} + 2] \frac{3}{n} = 15$
D $lim_{n \to \infty} \sum_{k = 1}^{n} [(\frac{3 k}{n})^{2} + 2] \frac{3}{n} = 5$

Solution & Step-by-step Explanation

The limit of the sum can be converted to a definite integral.Let \frac{k}{n} = Δ x, then \frac{3 k}{n} can be seen as x if we define the interval carefully.Alternatively, let x = \frac{3 k}{n} . Then d x = \frac{3}{n} .As k = 1, x \to 0 . As k = n, x \to 3 .The sum becomes: \int_{0}^{3} (x^{2} + 2) d x = [\frac{x ^{3}}{3} + 2 x]_{0}^{3} = (\frac{27}{3} + 6) - 0 = 9 + 6 = 15 The value of the limit is 15.

Try it yourself before checking the explanation above.

If lim_{n \to \infty} \sum_{k = 1}^{n} [(\frac{3 k}{n})^{2} + 2] \frac{3}{n} = \int_{0}^{6} f (x) d x, then:

f (0) = 1

f (x) = \frac{1}{3} x^{2} - \frac{1}{2} x + 1

lim_{n \to \infty} \sum_{k = 1}^{n} [(\frac{3 k}{n})^{2} + 2] \frac{3}{n} = 15

lim_{n \to \infty} \sum_{k = 1}^{n} [(\frac{3 k}{n})^{2} + 2] \frac{3}{n} = 5

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