Definite Integration Question - None

Question

By Leibnitz Theorem, if $\dfrac{d}{dx}\int_{0}^{x^3} k(t)\,dt = \dfrac{d}{dx}\left(x^{1+x^2}\right)$, find the value of $3k(1)$.

Accepted Answer

Step 1: Apply Leibniz Rule (Fundamental Theorem variant)

By Leibniz Theorem: d/dx∫₀^(x³) k(t)dt = k(x³)·d/dx(x³) = k(x³)·3x²

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

Let y = x^(1+x²). Taking ln: ln(y) = (1+x²)ln(x)

Differentiating: (1/y)·dy/dx = 2x·ln(x) + (1+x²)·(1/x)

So: dy/dx = x^(1+x²)[2x·ln(x) + (1+x²)/x]

Step 3: Equate both sides

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

Step 4: Find k(1)

Substituting x³ = 1 ⟹ x = 1:

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

k(1)·3 = 1·[0 + 2]

k(1) = 2/3

Step 5: Calculate 3k(1)

3k(1) = 3·(2/3) = 2

∴ Answer: B (2)

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