Definite Integration
Integration of periodic and absolute value functions
Grade 12
Question:
<p>Evaluate <math>\int_0^{np+w} |\sin x| \, dx</math>, where <math>n \in \mathbb{N}</math> and <math>0 \leq w < \pi</math></p>
<p>(a) <math>2n + \frac{1}{2}f(x)dx</math></p>
<p>(b) <math>2n + \cos w</math></p>
<p>(c) <math>(2n+1) - \cos w</math></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use periodicity of absolute value of sine function and split the integral. The integral over each complete period [0,π] equals 2.
<p><strong>Solution:</strong></p><p>Let <math>I = \int_0^{np+w} |\sin x| \, dx = \int_0^w |\sin x| \, dx + \int_w^{np+w} |\sin x| \, dx = I_1 + I_2</math></p><p><strong>For <math>I_1:</math></strong> <math>I_1 = \int_0^w \sin x \, dx</math> (since <math>0 \leq w < \pi</math> and <math>\sin x \geq 0</math> on <math>[0, \pi]</math>)</p><p><math>I_1 = [-\cos x]_0^w = -\cos w + 1 = 1 - \cos w</math></p><p><strong>For <math>I_2:</math></strong> <math>I_2 = \int_w^{np+w} |\sin x| \, dx = n \int_0^{\pi} |\sin x| \, dx</math></p><p><math>I_2 = n \int_0^{\pi} \sin x \, dx = n[-\cos x]_0^{\pi} = n(1+1) = 2n</math></p><p><strong>Therefore:</strong> <math>I = 1 - \cos w + 2n = (2n+1) - \cos w</math></p><p>∴ Answer is (c)</p>
Correct Answer: C