Question

The difference between the maximum and minimum value of the function \[ f(x) = \int_{0}^{x} (t^2 + t + 1)\, dt \] on the interval \[ [2, 3] \] is: 

• A. \( \frac{39}{6} \)
• B. \( \frac{49}{6} \)
• C. \( \frac{59}{6} \)
• D. \( \frac{69}{6} \)
• E. \( \frac{79}{6} \)

Hint:

If \( f'(x) > 0 \) on an interval, the function is increasing, so minimum occurs at left endpoint and maximum at right endpoint.

Answer 1

The Correct Option is C

Concept: To find maximum and minimum of a function on a closed interval, evaluate: - Critical points (where \( f'(x)=0 \)) - Endpoints of the interval

Step 1: Differentiate the function.
Given: \[ f(x) = \int_{0}^{x} (t^2 + t + 1)\,dt \] By Fundamental Theorem of Calculus: \[ f'(x) = x^2 + x + 1 \]

Step 2: Check critical points.
\[ x^2 + x + 1 = 0 \] Discriminant: \[ D = 1 - 4 = -3 < 0 \] So, no real roots ⇒ no critical points.

Step 3: Determine behavior.
Since \( x^2 + x + 1 > 0 \) for all \( x \), \[ f'(x) > 0 \Rightarrow f(x) \text{ is increasing on } [2,3] \]

Step 4: Evaluate at endpoints.
\[ f(x) = \int_{0}^{x} (t^2 + t + 1)\,dt = \left[\frac{t^3}{3} + \frac{t^2}{2} + t \right]_0^x \] \[ = \frac{x^3}{3} + \frac{x^2}{2} + x \] At \( x=2 \): \[ f(2) = \frac{8}{3} + 2 + 2 = \frac{8}{3} + 4 = \frac{20}{3} \] At \( x=3 \): \[ f(3) = \frac{27}{3} + \frac{9}{2} + 3 = 9 + \frac{9}{2} + 3 = \frac{33}{2} \]

Step 5: Find difference.
\[ \text{Difference} = f(3) - f(2) \] \[ = \frac{33}{2} - \frac{20}{3} = \frac{99 - 40}{6} = \frac{59}{6} \]

Step 6: Final answer.
\[ \boxed{\frac{59}{6}} \]