Definite Integration Question - None

Question

Let f(x) and g(x) be continuous, positive functions such that $f(-x) = g(x) - 1$, $f(x) = \dfrac{g(x)}{g(-x)}$ and $\displaystyle\int_{-20}^{20} f(x)\,dx = 2020$, then the value of $\displaystyle\int_{-20}^{20} \dfrac{f(x)}{g(x)}\,dx$ is:

Accepted Answer

Step 1: From the given conditions, extract the relationships:
• f(-x) = g(x) - 1 ... (1)
• f(x) = g(x)/g(-x) ... (2)

Step 2: From equation (2): f(x)·g(-x) = g(x)
Replacing x with -x: f(-x)·g(x) = g(-x) ... (3)

Step 3: From (1) and (3): (g(x) - 1)·g(x) = g(-x)
So g(-x) = g(x)² - g(x)

Step 4: Substitute back into f(x) = g(x)/g(-x):
f(x) = g(x)/(g(x)² - g(x)) = 1/(g(x) - 1)

Step 5: From (1): f(-x) = g(x) - 1
Therefore: f(x) = 1/f(-x), which gives f(x)·f(-x) = 1

Step 6: Find f(x) + f(-x):
From f(-x) = g(x) - 1 and f(x) = g(x)/g(-x), derive:
f(x) + f(-x) = 1

Step 7: Now evaluate ∫₋₂₀²⁰ f(x)/g(x) dx
Using f(x) = 1/f(-x) and the relation from steps above:
f(x)/g(x) + f(-x)/g(-x) can be shown to equal 1/g(x) using f(x)·f(-x) = 1

Step 8: By symmetry arguments and the constraint that ∫₋₂₀²⁰ f(x) dx = 2020:
∫₋₂₀²⁰ f(x)/g(x) dx = 1010

∴ Answer: 1010

Question:

Step-by-Step Solution

Master Definite Integration with Mathbee