Let f x be a differentiable function such that ∫ t t 2 x f x d x = 4 3 t 3 - 4 t 3 ∀ t ≥ 0 , then f 1 is equal to

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Question: Answered & Verified by Expert

Let

f (x)

be a differentiable function such that

\int_{t}^{t^{2}} x f (x) d x = \frac{4}{3} t^{3} - \frac{4 t}{3} \forall t \geq 0,

then

f (1)

is equal to

MathematicsDefinite IntegrationJEE Main

Options:

A $4$
B $\frac{4}{3}$
C $3$
D $\frac{8}{3}$

Solution:

2489 Upvotes Verified Answer

The correct answer is:

\frac{8}{3}

Differentiating both side using Leibnitz rule, we get,

t^{2} f (t^{2}) \cdot 2 t - t f (t) = 4 t^{2} - \frac{4}{3}

Substituting

t = 1

, we get,

f (1) = \frac{8}{3}

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