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JEE Main 2019 · 10 Apr · Shift I

JEE Main 2019 Mathematics question on Definite Integrals and Areas

lim⁡n→∞((n+1)1/3n4/3+(n+2)1/3n4/3+.....+(2n)1/3n4/3)\lim_{n \to \infty} \left( \frac{(n+1)^{1/3}}{n^{4/3}} + \frac{(n+2)^{1/3}}{n^{4/3}} + ..... + \frac{(2n)^{1/3}}{n^{4/3}} \right) is equal to
  1. A.
    43(2)4/3\frac{4}{3}(2)^{4/3}
  2. B.
    34(2)4/3−43\frac{3}{4}(2)^{4/3} - \frac{4}{3}
  3. C.
    34(2)4/3−34\frac{3}{4}(2)^{4/3} - \frac{3}{4}
  4. D.
    43(2)3/4\frac{4}{3}(2)^{3/4}
Show answer and solution

Correct option: C

Solution

The sum equals Σ(r=1 to n) of (n+r)^(1/3) / n^(4/3) = Σ(r=1 to n) (1/n)·(1 + r/n)^(1/3), which is a Riemann sum for ∫₀¹ (1+x)^(1/3) dx.

Evaluating: ∫₀¹ (1+x)^(1/3) dx = [(1+x)^(4/3) / (4/3)]₀¹ = (3/4)[(2)^(4/3) - 1] = (3/4)·2^(4/3) - 3/4.

[JEE Main 2019 · 10 Apr · Shift I]

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