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Find the area of the region bounded by the graph of f(x) = sin x and the x-axis on the interval [-pi/3, 2pi/3].

User Strayer
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1 Answer

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The area of the region bounded by the graph of
\(f(x) = \sin x\) and the x-axis on the interval
\([- (\pi)/(3), (2\pi)/(3)]\) is indeed 1 square unit.

To find the area of the region bounded by the graph of
\(f(x) = \sin x\) and the x-axis on the interval
\([- (\pi)/(3), (2\pi)/(3)]\), we can visualize the region by sketching the graph of
\(f(x) = \sin x\) within this interval.

The graph of
\(y = \sin x\) is a periodic function that oscillates between -1 and 1 as
\(x\) varies. The given interval
\([- (\pi)/(3), (2\pi)/(3)]\) captures a portion of this oscillation.

Let's visualize the graph:

- At
\(x = -(\pi)/(3)\), \(\sin x\) is positive, and the graph rises.

- At
\(x = 0\), the function crosses the x-axis.

- At
\(x = (\pi)/(3)\), \(\sin x\) is again positive, and the graph rises once more.

- At
\(x = (2\pi)/(3)\), the function crosses the x-axis again.

Now, within this interval, the shaded region represents the area between the curve
\(y = \sin x\) and the x-axis. To find this area, we set up the definite integral:


\[ \int_{-(\pi)/(3)}^{(2\pi)/(3)} \sin x \, dx \]

The integral of
\(\sin x\) is
\(-\cos x\), so applying the Fundamental Theorem of Calculus, we evaluate:


\[ \left[ -\cos x \right]_{-(\pi)/(3)}^{(2\pi)/(3)} \]

Plugging in the upper and lower limits:


\[ -\cos\left((2\pi)/(3)\right) + \cos\left(-(\pi)/(3)\right) \]

Evaluating these cosine values:


\[ -\left(-(1)/(2)\right) + (1)/(2) \]

Simplifying:


\[ (1)/(2) + (1)/(2) \]

Which equals 1.

User Henrycharles
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8.2k points