Answer:
Explanation:
To evaluate the integral of f(x) from -2 to 2, we need to split the integral into two parts based on the given piecewise function.
∫[-2, 2] f(x) dx = ∫[-2, 0] f(x) dx + ∫[0, 2] f(x) dx
For the interval -2 ≤ x ≤ 0:
f(x) = 2/(4 - x^2)
Let's calculate the integral on this interval:
∫[-2, 0] f(x) dx = ∫[-2, 0] (2/(4 - x^2)) dx
To simplify the integral, we can use a substitution. Let u = 4 - x^2, then du = -2x dx. Rearranging, dx = -du/(2x).
Substituting the limits and integrating:
∫[-2, 0] (2/(4 - x^2)) dx = ∫[4, 2] (1/u) (-du/(-2x)) = ∫[4, 2] (1/u) (du/(2x))
Now, we need to express the integral in terms of u. For the limits, when x = -2, u = 4 - (-2)^2 = 4 - 4 = 0, and when x = 0, u = 4 - 0^2 = 4.
∫[-2, 0] f(x) dx = ∫[4, 2] (1/u) (du/(2x)) = (1/2) ∫[4, 2] (du/u)
Evaluating the integral of (1/u) with respect to u:
(1/2) ∫[4, 2] (du/u) = (1/2) ln|u| ∣[4, 2] = (1/2) ln|2| - (1/2) ln|4|
For the interval 0 ≤ x ≤ 2:
f(x) = 0
Therefore, the integral on this interval is simply 0.
Combining the two parts, we have:
∫[-2, 2] f(x) dx = ∫[-2, 0] f(x) dx + ∫[0, 2] f(x) dx = (1/2) ln|2| - (1/2) ln|4| + 0
Simplifying:
∫[-2, 2] f(x) dx = (1/2) ln(2/4)
Therefore, the value of the integral is (1/2) ln(1/2), which is approximately -0.3466.