To solve the integral of the absolute value of the sine function over the given interval, we need to consider the behavior of the sine function within the interval. The sine function is positive in the first quadrant and negative in the third quadrant. Therefore, we need to split the integral at the points where the sine function changes sign, which are at \( \pi \) and \( 2\pi \).
The integral to be evaluated is \( \int_{0}^{3 \pi / 2} |\sin x| \, dx \). The sine function, \( \sin x \), is positive in the interval \( [0, \pi] \) and negative in the interval \( [\pi, 3\pi/2] \). Therefore, we need to split the integral at \( \pi \).
We split the integral into two parts:
\[
\int_{0}^{3 \pi / 2} |\sin x| \, dx = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{3 \pi / 2} -\sin x \, dx
\]
Evaluate the first integral:
\[
\int_{0}^{\pi} \sin x \, dx = \left[ -\cos x \right]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2
\]
Evaluate the second integral:
\[
\int_{\pi}^{3 \pi / 2} -\sin x \, dx = \left[ \cos x \right]_{\pi}^{3 \pi / 2} = \cos(3 \pi / 2) - \cos(\pi) = 0 - (-1) = 1
\]
Add the results of the two integrals:
\[
2 + 1 = 3
\]
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