Problem 39 - Calculus of a Single Variable

Question

Finding the Area of a Region In Exercises \(37-42,\) sketch the region bounded by the graphs of the functions and find the area of the region. $$ f(x)=2 \sin x, g(x)=\tan x, \quad-\frac{\pi}{3} \leq x \leq \frac{\pi}{3} $$

Step-by-Step Solution

Answer

The area of the region bounded by the two curves \(f(x)=2sin(x)\) and \(g(x)=tan(x)\) in the interval \(-pi/3 <= x <= pi/3\) is \(2/3\) square units.

Step 1: Identify the Interval of Intersection

Find the x-values where the functions intersect. Solve \(2sin(x) = tan(x)\) for the given range \(-\pi/3 <= x <= \pi/3\). The solutions are \(x = 0\) and \(x = pi/3\).

Step 2: Sketch the Curves and Identify the Area

From the graph, it can be observed that the upper curve is \(f(x)=2sin(x)\) and the lower curve is \(g(x)=tan(x)\) in the interval \([0, \pi/3]\), and vice-versa in the interval \([-pi/3, 0]\). The area under the curve from \(-pi/3\) to \(0\) is the same as the area from \(0\) to \(pi/3\). Thus, we can find the area from \(-pi/3\) to \(0\) and multiply it by 2.

Step 3: Set Up and Evaluate the Integral

The area A between the curves can be found using the formula: \(A = 2\int_{-pi/3}^0 [f(x)-g(x)] dx = 2\int_{-pi/3}^0 [2sin(x) - tan(x)] dx\). We use the properties of integrals to evaluate this expression. The result is \(2/3\).

Key Concepts

Integral Calculus

Integral calculus is a major branch of calculus focused on the accumulation of quantities and the spaces between curves. This is useful when determining areas under, above, or between curves. When you learned about derivatives, you found out how a quantity changes; now, with integrals, it's about summing up small pieces. In this problem, integral calculus assists us in identifying the area between parts of trigonometric functions.

• Integral calculus helps in computing total areas, volumes, and other quantities.
• Used in a variety of fields from physics to engineering and economics.

The area can be determined using definite integrals which sum the function's values over an interval. Functions need to be correctly set up in the integral to compute the desired area.

Trigonometric Functions

Trigonometric functions like sine and tangent are mathematical functions that relate angles of triangles to the lengths of the sides of triangles. In this instance, we're looking at the sine function and the tangent function. Each has unique properties that can be visualized graphically to find intersections or differences.

• Sine curve is smooth, periodic, and oscillates between -2 and 2 when multiplied.
• Tangent curve, however, has vertical asymptotes where it approaches infinity, notable near odd multiples of \(\frac{\pi}{2}\).

Identifying the intersection of these functions, solving \(2\sin x = \tan x\) over the interval helps us to set limits correctly for our integrals.

Area Between Curves

Finding the area between curves involves calculating the integral of the top function minus the bottom function. This is visualized by sketching the curves and determining where one function overtakes the other.
In this example:

• From \(-\frac{\pi}{3}\) to \(0\), the tangent function is below the sine function.
• From \(0\) to \(\frac{\pi}{3}\), roles reverse but due to symmetry, the areas are equal.

To find the overall area in symmetric intervals, calculate one and mirror it. Hence, integrating over \([-\frac{\pi}{3}, 0]\) and multiplying by 2 gives the complete area between the curves.

Definite Integrals

Definite integrals compute the net area under a curve from point \(a\) to point \(b\). They give us a precise measure of the area, fundamental in solving calculus problems involving areas.
In definite integrals:

• The limits of integration, like \([-\frac{\pi}{3}, 0]\), specify the beginning and end of the interval.
• Value is determined by the difference between the antiderivative at \(b\) and at \(a\).

By setting \(\int_{-\frac{\pi}{3}}^{0} [2\sin(x) - \tan(x)] dx\), the integral is evaluated, leading to a result of \(\frac{2}{3}\). Multiplying by 2 addresses the symmetry to find total area.