JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Definite Integrals

Let

[t]

denote the greatest integer

\leq t

. Then the value of

8 \cdot \int_{- \frac{1}{2}}^{1} ([2 x] + ∣ x ∣) d x

is ______.

Answer: 5

Solution

[- \frac{1}{2}, 0)

[2 x] = - 1

and

∣ x ∣ = - x

, so the integrand is

- 1 - x

; on

[0, \frac{1}{2})

[2 x] = 0

and it is

x

; on

[\frac{1}{2}, 1]

[2 x] = 1

and it is

1 + x

. Then

\int_{- 1/2}^{0} (- 1 - x) d x = - \frac{3}{8}

\int_{0}^{1/2} x d x = \frac{1}{8}

, and

\int_{1/2}^{1} (1 + x) d x = \frac{7}{8}

. The total is

- \frac{3}{8} + \frac{1}{8} + \frac{7}{8} = \frac{5}{8}

, so

8 \times \frac{5}{8} = 5

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