Math  /  Calculus

Questionπ2πsin3xdx\int_{-\pi}^{2 \pi} \sin 3 x d x

Studdy Solution

STEP 1

1. The integral to be solved is a definite integral of the function sin3x\sin 3x over the interval [π,2π][- \pi, 2\pi].
2. The antiderivative of sin3x\sin 3x will be used to evaluate the definite integral.

STEP 2

1. Find the antiderivative of sin3x\sin 3x.
2. Evaluate the definite integral using the Fundamental Theorem of Calculus.

STEP 3

Find the antiderivative of sin3x\sin 3x.
The antiderivative of sinkx\sin kx is 1kcoskx-\frac{1}{k} \cos kx. Therefore, the antiderivative of sin3x\sin 3x is:
sin3xdx=13cos3x+C\int \sin 3x \, dx = -\frac{1}{3} \cos 3x + C
where CC is the constant of integration.

STEP 4

Evaluate the definite integral using the Fundamental Theorem of Calculus.
The definite integral is given by:
π2πsin3xdx=[13cos3x]π2π\int_{-\pi}^{2\pi} \sin 3x \, dx = \left[ -\frac{1}{3} \cos 3x \right]_{-\pi}^{2\pi}
Calculate the value at the upper limit 2π2\pi:
13cos(3×2π)=13cos(6π)-\frac{1}{3} \cos(3 \times 2\pi) = -\frac{1}{3} \cos(6\pi)
Since cos(6π)=1\cos(6\pi) = 1, this becomes:
13×1=13-\frac{1}{3} \times 1 = -\frac{1}{3}
Calculate the value at the lower limit π-\pi:
13cos(3×π)=13cos(3π)-\frac{1}{3} \cos(3 \times -\pi) = -\frac{1}{3} \cos(-3\pi)
Since cos(3π)=1\cos(-3\pi) = -1, this becomes:
13×(1)=13-\frac{1}{3} \times (-1) = \frac{1}{3}
Now, evaluate the definite integral:
[13][13]=1313=23\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:
23\boxed{-\frac{2}{3}}

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