The curve y = x² – 7x + 10 intersects the x-axis at the points A and B. Then the area bounded by the curve and the line AB is

$4\frac{1}{2}$ sq. units

The curve y = x² – 7x + 10 intersects the x-axis at the points A(2, 0) and B(5, 0).

Hence the required area

$=-\int\limits_2^5y\,dx=-\int\limits_2^5(x^2-7x+10)dx$

$=-\begin{bmatrix}\frac{x^3}{3}-\frac{7x^2}{2}+10x\end{bmatrix}_2^5=4.5$

Hence (A) is the correct answer.