Question

\(6 \int_{0}^{\pi} (\sin 3x + \sin 2x + \sin x)dx\) is equal to:

Hint:

A useful property for sine integrals over \([0, \pi]\) is \(\int_0^\pi \sin(nx) dx\).
If `n` is an even integer, the integral is 0. If `n` is an odd integer, the integral is \(2/n\).
Using this: \(\int_0^\pi \sin(3x)dx = 2/3\), \(\int_0^\pi \sin(2x)dx = 0\), \(\int_0^\pi \sin(x)dx = 2\).
The sum is \(2/3 + 0 + 2 = 8/3\). Multiplying by 6 gives 16. This shortcut is much faster.

Answer 1

Correct Answer: 40

Step 1: Understanding the Question:
We are required to compute a definite integral of a sum of trigonometric functions over the interval from 0 to \(\pi\), and then multiply the result by 6.
Step 2: Key Formula or Approach:
The fundamental theorem of calculus will be used. The key integration formula needed is:
\[ \int \sin(ax) dx = -\frac{1}{a}\cos(ax) + C.
\] We will integrate each term of the sum individually.
Step 3: Detailed Explanation:
Let's first find the value of the integral part, \(I = \int_{0}^{\pi} (\sin 3x + \sin 2x + \sin x)dx\).
Using the integration formula for each term:
\[ I = \left[ -\frac{\cos(3x)}{3} - \frac{\cos(2x)}{2} - \cos(x) \right]_{0}^{\pi}.
\] Now, we evaluate the expression at the upper limit \(x=\pi\) and subtract the value at the lower limit \(x=0\).
Value at \(x=\pi\):
\[ \left( -\frac{\cos(3\pi)}{3} - \frac{\cos(2\pi)}{2} - \cos(\pi) \right) = \left( -\frac{-1}{3} - \frac{1}{2} - (-1) \right) = \frac{1}{3} - \frac{1}{2} + 1.
\] Value at \(x=0\):
\[ \left( -\frac{\cos(0)}{3} - \frac{\cos(0)}{2} - \cos(0) \right) = \left( -\frac{1}{3} - \frac{1}{2} - 1 \right).
\] Subtracting the value at 0 from the value at \(\pi\):
\[ I = \left(\frac{1}{3} - \frac{1}{2} + 1\right) - \left(-\frac{1}{3} - \frac{1}{2} - 1\right)
\] \[ I = \frac{1}{3} - \frac{1}{2} + 1 + \frac{1}{3} + \frac{1}{2} + 1 = \frac{2}{3} + 2 = \frac{8}{3}.
\] The problem asks for the value of \(6 \times I\).
\[ 6I = 6 \times \frac{8}{3} = 16.
\] Step 4: Final Answer:
The value of the expression is 16.