Math Problem Statement

intengrand between 3pi/2 and 0 of absolute value sin x

Solution

To find the integral of the absolute value of sin(x)\sin(x) from 00 to 3π2\frac{3\pi}{2}, we need to split the integral into intervals where sin(x)\sin(x) maintains a consistent sign, as the absolute value affects the function differently based on whether the sine function is positive or negative.

Step-by-Step Solution

  1. Identify intervals where sin(x)\sin(x) is positive or negative between 00 and 3π2\frac{3\pi}{2}:

    • From 00 to π\pi, sin(x)0\sin(x) \geq 0.
    • From π\pi to 3π2\frac{3\pi}{2}, sin(x)0\sin(x) \leq 0.
  2. Set up the integrals over these intervals:

    • For 0xπ0 \leq x \leq \pi, sin(x)=sin(x)|\sin(x)| = \sin(x).
    • For πx3π2\pi \leq x \leq \frac{3\pi}{2}, sin(x)=sin(x)|\sin(x)| = -\sin(x).

    Therefore, we split the original integral as: 03π2sin(x)dx=0πsin(x)dx+π3π2sin(x)dx\int_0^{\frac{3\pi}{2}} |\sin(x)| \, dx = \int_0^{\pi} \sin(x) \, dx + \int_{\pi}^{\frac{3\pi}{2}} -\sin(x) \, dx

  3. Evaluate each integral:

    • For the first integral: 0πsin(x)dx=[cos(x)]0π=cos(π)(cos(0))=(1)(1)=2\int_0^{\pi} \sin(x) \, dx = \left[-\cos(x)\right]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 2

    • For the second integral: π3π2sin(x)dx=[cos(x)]π3π2=(cos(3π2)cos(π))\int_{\pi}^{\frac{3\pi}{2}} -\sin(x) \, dx = -\left[\cos(x)\right]_{\pi}^{\frac{3\pi}{2}} = -\left(\cos\left(\frac{3\pi}{2}\right) - \cos(\pi)\right) Since cos(3π2)=0\cos\left(\frac{3\pi}{2}\right) = 0 and cos(π)=1\cos(\pi) = -1, (0(1))=1-\left(0 - (-1)\right) = -1

  4. Add the results of the integrals: 03π2sin(x)dx=2+1=3\int_0^{\frac{3\pi}{2}} |\sin(x)| \, dx = 2 + 1 = 3

Therefore, the integral of sin(x)|\sin(x)| from 00 to 3π2\frac{3\pi}{2} is 33.

Would you like more details or have any questions about this solution?

Relative Questions

  1. How would the integral change if the limits were from 00 to 2π2\pi?
  2. What is the general strategy for integrating the absolute value of a trigonometric function?
  3. How can we integrate other functions involving absolute values?
  4. What role does symmetry play in integrals involving trigonometric functions?
  5. How would this integral differ if it were 03π2cos(x)dx\int_0^{\frac{3\pi}{2}} |\cos(x)| \, dx?

Tip

When integrating absolute values of trigonometric functions, always identify the intervals where the function is positive or negative to simplify the absolute value expressions.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Trigonometric Functions
Absolute Value

Formulas

Integral of sin(x): ∫ sin(x) dx = -cos(x)
Integral of -sin(x): ∫ -sin(x) dx = cos(x)
Splitting integrals based on the sign of sin(x)

Theorems

Fundamental Theorem of Calculus
Properties of the sine function on different intervals

Suitable Grade Level

College-level Calculus