JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Definite Integrals

Let

f (x)

and

g (x)

be two functions satisfying

f (x^{2}) + g (4 - x) = 4 x^{3}

and

g (4 - x) + g (x) = 0

, the value of

\int_{- 4}^{4} f (x^{2}) d x

is ____.

Answer: 512

Solution

The second relation gives

g (4 - x) = - g (x)

, so the first becomes

f (x^{2}) = 4 x^{3} - g (4 - x) = 4 x^{3} + g (x)

Since

f (x^{2})

is even in

x

\int_{- 4}^{4} f (x^{2}) d x = 2 \int_{0}^{4} f (x^{2}) d x = 2 \int_{0}^{4} (4 x^{3} + g (x)) d x

The substitution

x \to 4 - x

turns

\int_{0}^{4} g (x) d x

into

\int_{0}^{4} g (4 - x) d x

, which equals

- \int_{0}^{4} g (x) d x

, so that integral is its own negative and vanishes. What remains is

2 \int_{0}^{4} 4 x^{3} d x = 2 [x^{4}]_{0}^{4} = 512

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Integer answer

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