find the value(s) of x for which ∫ₓ² f(t)dt = 0, where f(t) is an odd function. a) -2 only b) -2, 0, and 2…

find the value(s) of x for which ∫ₓ² f(t)dt = 0, where f(t) is an odd function. a) -2 only b) -2, 0, and 2 c) 0 only d) -2 and 2 only e) 2 only

find the value(s) of x for which ∫ₓ² f(t)dt = 0, where f(t) is an odd function. a) -2 only b) -2, 0, and 2 c) 0 only d) -2 and 2 only e) 2 only

Answer

Explanation:

Step1: Recall property of odd - function integral

For an odd function (y = f(t)), (\int_{-a}^{a}f(t)dt = 0). Also, (\int_{x}^{2}f(t)dt=0) implies (\int_{x}^{0}f(t)dt+\int_{0}^{2}f(t)dt = 0), or (\int_{x}^{0}f(t)dt=-\int_{0}^{2}f(t)dt). Since (f(t)) is odd, (\int_{-2}^{2}f(t)dt = 0). If (x=-2), (\int_{-2}^{2}f(t)dt = 0). Also, if (x = 2), (\int_{2}^{2}f(t)dt=0) (by the property (\int_{a}^{a}g(x)dx = 0) for any function (g(x))). And for (x = 0), (\int_{0}^{2}f(t)dt) is not necessarily (0) for a general odd - function (f(t)) (but (\int_{-2}^{2}f(t)dt = 0) and (\int_{2}^{2}f(t)dt = 0)).

Step2: Analyze the integral values

We know that (\int_{x}^{2}f(t)dt = 0). When (x=-2), (\int_{-2}^{2}f(t)dt = 0) due to the odd - function property. When (x = 2), (\int_{2}^{2}f(t)dt=0) by the basic integral property (\int_{a}^{a}f(x)dx = 0).

Answer:

D. -2 and 2 only