
Matthew S. answered 02/24/20
PhD in Mathematics with extensive experience teaching Calculus
Answer #4 is the correct one.
Since f is continuous on [0, √5], the mean value theorem (MVT) for integrals applies. According to MVT,
(I) ∫0√5f(x)dx = f(c) · (√5 - 0) for some c in the interval.
∫0√5(5 - x2) dx = 5x - x3/3 evaluated at √5 and 0. So the integral is 5√5 - 5√5/3, which simplifies to 10√5/3.
Solving equation (I) for f(c) gives you f(c) = 10/3. The problem statement asks for c. 10/3 = 5 - c2, so c2 = 15/3 - 10/3 or 5/3. Lastly, c = sqrt(5/3). After rationalizing, we arrive at c = √15/3