Question$\int_{-\pi}^{2 \pi} \sin 3 x d x$

Studdy Solution
Evaluate the definite integral using the Fundamental Theorem of Calculus.
The definite integral is given by:
$\int_{-\pi}^{2\pi} \sin 3x \, dx = \left[ -\frac{1}{3} \cos 3x \right]_{-\pi}^{2\pi}$
Calculate the value at the upper limit $2\pi$:
$-\frac{1}{3} \cos(3 \times 2\pi) = -\frac{1}{3} \cos(6\pi)$
Since $\cos(6\pi) = 1$, this becomes:
$-\frac{1}{3} \times 1 = -\frac{1}{3}$
Calculate the value at the lower limit $-\pi$:
$-\frac{1}{3} \cos(3 \times -\pi) = -\frac{1}{3} \cos(-3\pi)$
Since $\cos(-3\pi) = -1$, this becomes:
$-\frac{1}{3} \times (-1) = \frac{1}{3}$
Now, evaluate the definite integral:
$\left[ -\frac{1}{3} \right] - \left[ \frac{1}{3} \right] = -\frac{1}{3} - \frac{1}{3} = -\frac{2}{3}$
The value of the definite integral is:
$\boxed{-\frac{2}{3}}$