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Chapter 12: Problem 1011

\(2 \pi \int_{0}(\sin x+|\sin x|) d x\) is equal to \(\ldots \ldots\) (a) 0 (b) 2 (c) \(-2\) (d) 4

Short Answer

Expert verified
The expression \(2\pi \int_{0}(\sin x+|\sin x|) dx\) is equal to 4. The correct answer is (d) 4.

Step by step solution

01

Analyze functions within the integral

We must understand the behavior of sin(x) and its absolute value |sin(x)|. Sin(x) is a periodic function that oscillates between -1 and 1, and its period is 2π. However, the absolute value function |sin(x)| makes all negative values of sin(x) positive, effectively "folding" the negative half of the sine wave onto positive values while keeping the positive half unchanged.
02

Find where sin(x) is positive and negative

Within the interval [0, 2π], sin(x) is positive in the range (0, π) and negative in the range (π, 2π). Therefore, we can rewrite the expression as follows: \(\int_0^{2\pi}(\sin x + |\sin x|)dx = \int_0^\pi (\sin x + \sin x)dx + \int_\pi^{2\pi} (\sin x - \sin x)dx\)
03

Simplify the new expression

We can now simplify both expressions within the integral: \(\int_0^{2\pi}(\sin x + |\sin x|)dx = \int_0^\pi (2\sin x)dx + \int_\pi^{2\pi} (0)dx\)
04

Solve the integral separately

We will solve each integral separately: For the first integral: \(\int_0^\pi (2\sin x)dx = 2\int_0^\pi \sin x dx\) Now, the integral of sin(x) is -cos(x), so we have: \(2[-\cos x]_0^\pi = 2[-\cos(\pi) + \cos(0)] = 2[1 + 1] = 4\) For the second integral: \(\int_\pi^{2\pi} (0)dx = 0\)
05

Combine the integrals and choose the correct answer

We have found the values for each integral, so now we just need to put them back together: \(\int_0^{2\pi}(\sin x + |\sin x|)dx = 4 + 0 = 4\) Thus, the expression \(2\pi \int_{0}(\sin x+|\sin x|) dx\) is equal to 4. The correct answer is (d) 4.

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Most popular questions from this chapter

The value of integral \(\pi \int_{0}[\\{\sin (2 n+1)(x / 2)\\} /\\{\sin (x / 2)\\}] d x\) is...... (a) 0 (b) \((\pi / 2)\) (c) \(\pi\) (d) \(2 \pi\)

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