Let $F(x) = \int_{-1}^x \sqrt{4+t^2} dt$ and $G(x) = \int_x^1 \sqrt{4+t^2} dt$ then the value of $(FG)'(0)$ is____ (where dash denotes the derivative).
Step-by-Step Solution
Key Concept: Use the Leibniz rule for differentiation of products of integrals with variable limits: apply the chain rule and fundamental theorem of calculus.
Given $F(x) = \int_{-1}^x f(t)\, dt$ and $G(x) = \int_x^1 f(t)\, dt$ with $f(t) = \sqrt{4+t^2}$, compute $H(x) = F(x)G(x)$. By the product rule, $H'(x) = F'(x)G(x) + F(x)G'(x)$. Since $F'(x) = f(x) = \sqrt{4+x^2}$ and $G'(x) = -f(x) = -\sqrt{4+x^2}$, we have $H'(x) = \sqrt{4+x^2}\left(\int_x^1 f(t)\, dt\right) + \left(\int_{-1}^x f(t)\, dt\right)(-\sqrt{4+x^2}) = \sqrt{4+x^2}\left(\int_x^1 f(t)\, dt - \int_{-1}^x f(t)\, dt\right)$.
Correct Answer: 0