Definite Integration Question - None

Question

892. Let $f$ be a continuous and even function such that $\int_0^a f(x)\,dx = 10$. If $g(x)$ is a continuous positive function such that $g(x)g(-x) = 1$ and $\int_0^a g(x)\,dx = 5$, then find the value of $\displaystyle\int_{-a}^{a} \dfrac{f(x)}{1+g(x)}\,dx$.

Accepted Answer

Step 1: Since f is even: ∫₋ₐᵃ f(x)dx = 2∫₀ᵃ f(x)dx = 2(10) = 20

Step 2: From g(x)g(-x) = 1, we get g(-x) = 1/g(x). Therefore:

1/(1+g(x)) + 1/(1+g(-x)) = 1/(1+g(x)) + 1/(1+1/g(x)) = 1/(1+g(x)) + g(x)/(g(x)+1) = 1

Step 3: Split the integral:

∫₋ₐᵃ f(x)/(1+g(x))dx = ∫₋ₐ⁰ f(x)/(1+g(x))dx + ∫₀ᵃ f(x)/(1+g(x))dx

Step 4: For the first integral, substitute u = -x (noting f(-x) = f(x)):

∫₋ₐ⁰ f(x)/(1+g(x))dx = ∫₀ᵃ f(u)/(1+g(-u))du = ∫₀ᵃ f(u)g(u)/(1+g(u))du

Step 5: Add both integrals:

∫₋ₐᵃ f(x)/(1+g(x))dx = ∫₀ᵃ f(x)[1/(1+g(x)) + g(x)/(1+g(x))]dx = ∫₀ᵃ f(x)dx = 10

∴ Answer: 10

Question:

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