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JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Definite Integrals

The value of the integral

\int_{- 2}^{2} \frac{∣ x ^{3} + x ∣}{( e ^{x ∣ x ∣} + 1 )} d x

is equal to :

(A) $5 e^{2}$
(B) $3 e^{- 2}$
(C)4
(D)6✓ Correct

Answer: (D)

Solution

Let

f (x) = \frac{∣ x ^{3} + x ∣}{e ^{x ∣ x ∣} + 1}

. Check if

f (x) + f (- x)

simplifies:

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

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

= ∣ x^{3} + x ∣ (\frac{1}{e ^{x ∣ x ∣} + 1} + \frac{e ^{x ∣ x ∣}}{e ^{x ∣ x ∣} + 1}) = ∣ x^{3} + x ∣

For

x \in [0, 2]

,

∣ x^{3} + x ∣ = x^{3} + x

. Therefore:

\int_{- 2}^{2} f (x) d x = \int_{0}^{2} (x^{3} + x) d x = [\frac{x ^{4}}{4} + \frac{x ^{2}}{2}]_{0}^{2} = 4 + 2 = 6

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option

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