Question

The difference between the greatest and the least value of the function F(x) = $\int_{0}^{x}{(t + 1)}$dt on [2, 3] is –

• A. 3
• B. 2
• C. 7/2
• D. 3/2

Answer 1

Answer: (C)

7/2

Answer 2

Differentiating the given function, we get

F¢(x) = [t + 1]_{t = x} $\frac{dx}{dx}$ – [t + 1]_{t = 0} $\frac{dx}{dx}$ = x + 1.

This is positive for all x Ī [2, 3], so F is an increasing function in this interval. Therefore its greatest value is

F(3) = $\int_{0}^{3}{(t + 1)}$dt and its least value is F(2) = $\int_{0}^{2}{(t + 1)}$dt, so that the required difference between these values is

$\int_{0}^{3}{(t + 1)}$ dt – $\int_{0}^{2}{(t + 1)}$ dt = $\frac{7}{2}$.