JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Definite Integrals

The value of

\int_{- 1}^{1} x^{2} e^{[x^{3}]} d x

, where

[t]

denotes the greatest integer

\leq t

, is :

Answer: (B)

Solution

[- 1, 0)

the cube

x^{3}

lies in

[- 1, 0)

, so

[x^{3}] = - 1

, and on

[0, 1)

it lies in

[0, 1)

, so

[x^{3}] = 0

; the single point

x = 1

changes nothing in the integral. Hence

\int_{- 1}^{1} x^{2} e^{[x^{3}]} d x = \frac{1}{e} \int_{- 1}^{0} x^{2} d x + \int_{0}^{1} x^{2} d x = \frac{1}{3 e} + \frac{1}{3} .

Combining the two terms, the integral equals

\frac{e + 1}{3 e}

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option

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