JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Definite Integrals

\int_{- a}^{a} (∣ x ∣ + ∣ x - 2∣) d x = 22

(a > 2)

and

[x]

denotes the greatest integer

\leq x

, then

\int_{- a}^{a} (x + [x]) d x

is equal to ______.

Answer: -3

Solution

The first integral fixes

a

. Since

\int_{- a}^{a} ∣ x ∣ d x = a^{2}

and, for

a > 2

\int_{- a}^{a} ∣ x - 2∣ d x = a^{2} + 4

, their sum is

2 a^{2} + 4 = 22

, giving

a^{2} = 9

and

a = 3

. For the required integral,

\int_{- 3}^{3} x d x = 0

by symmetry, while the greatest-integer part gives, using

[x] + [- x] = - 1

at every non-integer point,

\int_{- 3}^{3} [x] d x = \int_{0}^{3} ([x] + [- x]) d x = \int_{0}^{3} (- 1) d x = - 3

. Adding the two parts gives

0 + (- 3) = - 3

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