The area (in sq. units) of the region bounded by the curve y = x^2-5x…

The correct answer is Option (1) → \(\displaystyle\frac{9}{2}\)

Since \(x^{2}-5x+4 = (x-1)(x-4)\), the curve meets the x-axis at x = 1 and x = 4 and lies below the x-axis in between, so the area is the absolute value of the integral:

    \( \displaystyle \int_{1}^{4}(x^{2}-5x+4)\,dx = \left[\frac{x^{3}}{3}-\frac{5x^{2}}{2}+4x\right]_{1}^{4} \)

    \( \displaystyle = \left(\frac{64}{3}-40+16\right)-\left(\frac{1}{3}-\frac{5}{2}+4\right) \)

    \( \displaystyle = -\frac{8}{3}-\frac{11}{6} = -\frac{9}{2} \)

So

    \( \displaystyle A = \left|-\frac{9}{2}\right| = \frac{9}{2} \)

Shortcut: The area between \(y = (x-\alpha)(x-\beta)\) and the x-axis is \(\frac{(\beta-\alpha)^{3}}{6}\). Here \(\frac{(4-1)^{3}}{6} = \frac{27}{6} = \frac{9}{2}\).