Let $f(x)=\int \mathrm{x}^{3} \sqrt{3-\mathrm{x}^{2}} \mathrm{dx}$. If $5 f(\sqrt{2})=-4$, then $f(1)$ is equal to

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Indefinite Integrals

Let

f (x) = \int x^{3} 3 - x^{2} dx

. If

5 f (2) = - 4

, then

f (1)

is equal to

Answer: (D)

Substituting

3 - x^{2} = t^{2}

gives

x d x = - t d t

and

x^{2} = 3 - t^{2}

, so

f (x) = \int (3 - t^{2}) t (- t d t) = \int (t^{4} - 3 t^{2}) d t = \frac{t ^{5}}{5} - t^{3} + c = \frac{( 3 - x ^{2} ) ^{5/2}}{5} - (3 - x^{2})^{3/2} + c .

x = 2

t = 1

, so

f (2) = \frac{1}{5} - 1 + c

; the condition

5 f (2) = - 4

gives

1 - 5 + 5 c = - 4

, hence

c = 0

. Then

f (1) = \frac{2 ^{5/2}}{5} - 2^{3/2} = 2 (\frac{4}{5} - 2) = - \frac{6 2}{5}

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option

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