Let g be a differentiable function satisfying $\int_\limits{0}^{x}$ (x - t + 1) g(t) dt = x 4 + x 2 for all $x \geq 0$ . The value of $\int_\limits{0}^{1} \frac{12}{g^{\prime}(x)+g(x)+10}$ dx is equal to :

Definite Integration

Grade 12

Question:

Let g be a differentiable function satisfying $\int_\limits{0}^{x}$ (x - t + 1) g(t) dt = x4 + x2 for all $x \geq 0$. The value of $\int_\limits{0}^{1} \frac{12}{g^{\prime}(x)+g(x)+10}$ dx is equal to :

$\frac{\pi}{3}$

$\frac{\pi}{2}$

$\frac{\pi}{4}$

$\frac{\pi}{6}$

Step-by-Step Solution

Key Concept: Differentiate the given integral equation twice using the Leibniz Rule to obtain a direct expression for g'(x) + g(x), which simplifies the integrand into a standard form.

$\frac{\pi}{4}$

Correct Answer: C

Question:

Step-by-Step Solution

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