The value of \(\int_{3}^{4} \sqrt{(4-x)(x-3)} d x\) is

Let \(I=\int_{3}^{4} \sqrt{(4-x)(x-3)} d x\)
\(=\int_{3}^{4} \sqrt{-x^{2}+7 x-12} d x\)
\(=\int_{3}^{4} \sqrt{-\left(x^{2}-7 x+\frac{49}{4}-\frac{49}{4}\right)-12} d x\)
\(=\int_{3}^{4} \sqrt{-\left(x-\frac{7}{2}\right)^{2}+\frac{49}{4}-12} d x\)
\(=\int_{3}^{4} \sqrt{\frac{1}{4}-\left(x-\frac{7}{2}\right)^{2}} d x\)
Let \(t=x-\frac{7}{2} \Rightarrow d t=d x\)
\(\therefore\) Upper limit \(=\frac{1}{2}\) and lower limit \(=-\frac{1}{2}\)
\(
\begin{array}{l}
=\int_{-1 / 2}^{1 / 2} \sqrt{\left(\frac{1}{2}\right)^{2}-t^{2}} d t \\
=2 \int_{0}^{1 / 2} \sqrt{\left(\frac{1}{2}\right)^{2}-t^{2}} d t \\
=2\left[\frac{t}{2} \sqrt{\frac{1}{4}-t^{2}}+\frac{1}{8} \sin ^{-1} 2 t\right]_{0}^{1 / 2} \\
=2\left[0+\frac{1}{8} \times \frac{\pi}{2}\right] \\
=\frac{\pi}{8}
\end{array}
\)