∫_0^3≤ft|x^2-3x+2|dx is equal to

The correct answer is Option (3) → \(\displaystyle\frac{11}{6}\)

Since \(x^{2}-3x+2 = (x-1)(x-2)\), it is positive on \([0, 1)\), negative on \((1, 2)\) and positive on \((2, 3]\). Splitting the interval at the roots, with \(F(x) = \frac{x^{3}}{3} - \frac{3x^{2}}{2} + 2x\):

    \( \displaystyle \int_{0}^{1}(x^{2}-3x+2)\,dx = F(1) - F(0) = \frac{5}{6} \)

    \( \displaystyle \int_{1}^{2}-(x^{2}-3x+2)\,dx = -\left(F(2) - F(1)\right) = -\left(\frac{2}{3} - \frac{5}{6}\right) = \frac{1}{6} \)

    \( \displaystyle \int_{2}^{3}(x^{2}-3x+2)\,dx = F(3) - F(2) = \frac{3}{2} - \frac{2}{3} = \frac{5}{6} \)

Adding:

    \( \displaystyle \frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6} \)

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{3}{2}\) — Incorrect. This is \(\int_{0}^{3}(x^{2}-3x+2)\,dx\) without the modulus; the part on (1, 2) is subtracted instead of added.

• Option (2) \(\displaystyle\frac{5}{6}\) — Incorrect. This is only the part over \([0, 1]\) (or over \([2, 3]\)).

• Option (4) \(\displaystyle\frac{7}{6}\) — Incorrect. The three parts add to \(\frac{11}{6}\), not \(\frac{7}{6}\).