Second Fundamental Theorem of Calculus

Question

Find the maximum value of F(x) = ∫ 0 to x (sin(t²)) dt on the interval 0 ≤ x ≤ 2.5.

Maximum point = ( _, _ )

Answer 1

F(x) = ∫_{(from 0 to x)} sin(t²)dt

F'(x) = sin(x²) > 0 when 0 < x² < π. So, 0 < x < √ π ≈ 1.77

F'(√ π) = sin π = 0

F'(x) = sin(x²) < 0 when π < x² < 2π. So, √ π < x < √ (2π) (1.77 < x < 2.5066)

So, F(x) is increasing on (0, 1.77) and is decreasing on (1.77, 2.5)

Max of F(x) on (0, 2.5) occurs when x = √π ≈ 1.77.