Let \(f(x) = \int_{0}^{x}\displaystyle(t^{2}-4t+3)\,dt\). Which of the following statements are correct?
(A) \(f'(2) = -1\)
(B) \(f(3) = 0\)
(C) f is decreasing on the interval (1, 3)
(D) \(f(1) = 1\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (C) only
By the first fundamental theorem of calculus,
\(f'(x) = x^{2}-4x+3 = (x-1)(x-3)\)
Integrating directly,
\( \displaystyle f(x) = \frac{x^{3}}{3} - 2x^{2} + 3x \)
Then
\(f'(2) = 4 - 8 + 3 = -1\)
\(f(3) = 9 - 18 + 9 = 0\)
\( \displaystyle f(1) = \frac{1}{3} - 2 + 3 = \frac{4}{3} \)
For \(1 \lt x \lt 3\), the factor \(x-1\) is positive and \(x-3\) is negative, so \(f'(x) \lt 0\).
So \(f'(2) = -1\), \(f(3) = 0\), \(f(1) = \frac{4}{3}\), and \(f'(x) \lt 0\) on (1, 3).
Checking the statements
(A) \(f'(2) = -1\) — Correct. \(f'(2) = 4 - 8 + 3 = -1\).
(B) \(f(3) = 0\) — Correct. \(f(3) = 9 - 18 + 9 = 0\).
(C) f is decreasing on the interval (1, 3) — Correct. \(f'(x) = (x-1)(x-3) \lt 0\) for \(1 \lt x \lt 3\).
(D) \(f(1) = 1\) — Incorrect. \(f(1) = \frac{1}{3} - 2 + 3 = \frac{4}{3}\).