JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Definite Integrals

Let

f (x) = \int_{0}^{x} g (t) d t

, where

g

is a non zero even function. If

f (x + 5) = g (x)

, then

\int_{0}^{x} f (t) d t

equals

Answer: (A)

Solution

Since

g

is even and

f (x) = \int_{0}^{x} g

, differentiating gives

f^{'} (x) = g (x)

, and a short check shows

f

is odd,

f (- x) = - f (x)

. The condition

f (x + 5) = g (x) = f^{'} (x)

together with the evenness of

g

forces

f (x + 5) = - f (x - 5)

, so

f (x) = - g (x + 5)

. Then

\int_{0}^{x} f (t) d t = - \int_{0}^{x} g (t + 5) d t

, and substituting

u = t + 5

turns this into

- \int_{5}^{x + 5} g (u) d u = \int_{x + 5}^{5} g (t) d t

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option

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