Define F(x) = integral from x to (x² + pi/6) of 2*cos²(t) dt for all real x. Let f: [0, 1/2] - [0, infinity) be continuous. If, for every a in [0, 1/2], the quantity F'(a) + 2 equals the area of the region bounded by x = 0, y = 0, the curve y = f(x) and the line x = a, then find f(0).

Question

Define F(x) = integral from x to (x² + pi/6) of 2*cos²(t) dt for all real x. Let f: [0, 1/2] -> [0, infinity) be continuous. If, for every a in [0, 1/2], the quantity F'(a) + 2 equals the area of the region bounded by x = 0, y = 0, the curve y = f(x) and the line x = a, then find f(0).

Accepted Answer

3. The stated area equals integral₀^a f(x) dx, so integral₀^a f = F'(a) + 2. Differentiating both sides with respect to a gives f(a) = F''(a). Hence f(0) = F''(0). Computing F'(x) by Leibniz rule and differentiating again, all terms with an x factor vanish at 0, leaving F''(0) = 4*cos²(pi/6) = 3.

Define F(x) = integral from x to (x² + pi/6) of 2*cos²(t) dt for all real x. Let f: [0, 1/2] -> [0, infinity) be continuous. If, for every a in [0, 1/2], the quantity F'(a) + 2 equals the area of the region bounded by x = 0, y = 0, the curve y = f(x) and the line x = a, then find f(0).

Solution

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