Tích phân \(I = \int\limits_{\frac{5}{2}}^3 {\sqrt {\left( {x - 1} \right)\left( {3 - x} \right)} dx} \) có giá trị là:

* Hướng dẫn giải

Ta có:

\(I = \int\limits_{\frac{5}{2}}^3 {\sqrt {\left( {x - 1} \right)\left( {3 - x} \right)} dx} = \int\limits_{\frac{5}{2}}^3 {\sqrt { - 3 - {x^2} + 2x} dx} = \int\limits_{\frac{5}{2}}^3 {\sqrt {1 - {{\left( {x - 2} \right)}^2}} dx} \).

Đặt \(x - 2 = \sin t,t \in \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right] \Rightarrow dx = \cos tdt\).

Đổi cận \(\left\{ \begin{array}{l} x = \frac{5}{2} \Rightarrow t = \frac{\pi }{6}\\ x = 3 \Rightarrow t = \frac{\pi }{2} \end{array} \right.\).

\( \Rightarrow I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\sqrt {1 - {{\sin }^2}t} .\cos tdt} \\= \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {{{\cos }^2}tdt} \\= \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{1 + \cos 2t}}{2}dt = \frac{1}{2}\left. {\left( {x + \frac{1}{2}\sin 2t} \right)} \right|_{\frac{\pi }{6}}^{\frac{\pi }{2}}} \\= \frac{\pi }{6} - \frac{{\sqrt 3 }}{8}\)