Question

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].

Answer 1

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.