Definite Integration Question - None

Question

888. Let $k(x)$ be a continuous function satisfying the equation $\displaystyle\int_0^{x^3} k(t)\, dt = x^{1+x^2}$, find the value of $3k(1)$.

Accepted Answer

Step 1: Differentiate both sides with respect to x using Leibniz rule.

Given: ∫₀^(x³) k(t)dt = x^(1+x²)

Left side: d/dx[∫₀^(x³) k(t)dt] = k(x³)·d/dx(x³) = k(x³)·3x²

Step 2: Differentiate the right side x^(1+x²).

Using logarithmic differentiation: d/dx[x^(1+x²)] = x^(1+x²)·d/dx[(1+x²)ln(x)]

= x^(1+x²)·[ln(x)·2x + (1+x²)·(1/x)]

= x^(1+x²)·[2x·ln(x) + (1+x²)/x]

Step 3: Equate both sides and substitute x=1.

k(x³)·3x² = x^(1+x²)·[2x·ln(x) + (1+x²)/x]

At x=1: k(1)·3(1)² = 1^(1+1)·[2(1)·ln(1) + (1+1)/1]

k(1)·3 = 1·[0 + 2]

3k(1) = 2

∴ Answer: 2

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