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(D) Given $f(x) = \int_{2}^{x} (2t - 5) \, dt$. Evaluating the integral: $f(x) = [t^2 - 5t]_{2}^{x} = (x^2 - 5x) - (4 - 10) = x^2 - 5x + 6$. The graph

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$\frac{\pi}{2}$

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(D) Given $f(x) = \int_{2}^{x} (2t - 5) \, dt$. Evaluating the integral: $f(x) = [t^2 - 5t]_{2}^{x} = (x^2 - 5x) - (4 - 10) = x^2 - 5x + 6$. The graph cuts the $x$-axis where $f(x) = 0$. $x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0$,so $x = 2$ and $x = 3$. The points of intersection are $(2, 0)$ and $(3, 0)$. The slope of the tangent is $f'(x) = \frac{d}{dx}(x^2 - 5x + 6) = 2x - 5$. At $x = 2$,$m_1 = 2(2) - 5 = -1$. At $x = 3$,$m_2 = 2(3) - 5 = 1$. The angle $	heta$ between the tangents is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$. $	an 	heta = \left| \frac{-1 - 1}{1 + (-1)(1)} ight| = \left| \frac{-2}{0} ight| = \infty$. Thus,$	heta = \frac{\pi}{2}$.

The angle between the tangent lines to the graph of the function $f(x) = \int_{2}^{x} (2t - 5) \, dt$ at the points where the graph cuts the $x$-axis is

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