Mark M. answered 08/23/15
Tutor
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Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
FALSE:
Let f(x) = x, g(x) = x2 where 0 ≤ x ≤ 1
Then ∫f(x)dx from x = 0 to x = 1 is 1/2
∫g(x)dx from x = 0 to x = 1 is 1/3
f(x)g(x) = x3 So, ∫[f(x)g(x) from x = 0 to x = 1 is 1/4
But, ∫f(x)dx ∫g(x)dx from x = 0 to x = 1 is (1/2)(1/3) = 1/6
So, ∫[f(x)g(x)]dx ≠ ∫f(x)dx ∫g(x)dx