Question

Definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand. show the region in question, and interpret your result. $$\int_{-1}^{3} \sqrt{4-(x-1)^{2}} d x$$

Short answer

Question: Evaluate the definite integral of the given function using geometry: \(\int_{-1}^{3} \sqrt{4-(x-1)^{2}} dx\) Answer: The definite integral of the given function between x=-1 and x=3 is equal to the area of the region between the semicircle and the x-axis, which is \(\frac{2}{3}\pi\).

Step-by-step solution

Identify the Shape

The given function is \(\int_{-1}^{3} \sqrt{4-(x-1)^{2}} dx\). We can observe that the function has the equation of a semicircle with radius 2, centered at (1, 0): $$4 = (x-1)^2 + y^2$$ $$y = \sqrt{4 - (x-1)^2}$$ The required region lies between x=-1 and x=3.

Sketch the Graph and Identify the Region

Now, let's sketch the graph of the function which is a semicircle with radius 2 and centered at (1, 0). The graph will form an upper semicircle that starts at x=-1 and ends at x=3. Draw the region enclosed by the semicircle and the x-axis between x=-1 and x=3.

Calculate the Area Using Geometry

The semicircle has a radius of 2. The area of a full circle with radius 2 is given by the formula: $$A_{circle} = \pi r^2 = \pi (2)^2 = 4\pi$$ However, since we are given an upper semicircle, we only want half the area of this circle: $$A_{semicircle} = \frac{1}{2} A_{circle} = \frac{1}{2}(4\pi) = 2\pi$$ Now, we need to find the area of the region between x=-1 and x=3. This region is also a sector, with a central angle of 120 degrees (since the arc spans from x=-1 to x=3 and the distance between these points on the x-axis is 4 units, and a semicircle arc spans a total of 180 degrees). The area of the sector is given by the formula: $$A_{sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} (2)^2 \left(\frac{120}{180}\pi\right) = \frac{4}{3}\pi$$ Finally, we need to subtract the area of the sector from the area of the semicircle to obtain the area of the region covered by the definite integral: $$A_{region} = A_{semicircle} - A_{sector} = 2\pi - \frac{4}{3}\pi = \frac{2}{3}\pi$$

Interpret the Result

The definite integral of the given function between x=-1 and x=3 is equal to the area of the region between the semicircle and the x-axis, which is: $$\int_{-1}^{3} \sqrt{4-(x-1)^{2}} dx = \frac{2}{3}\pi$$

Key concepts

Geometric Interpretation of Definite Integrals

Definite integrals represent a fundamental concept in calculus, often associated with the geometric idea of finding the area under a curve. When you are given a function and the limits of integration, calculating the definite integral can be visualized as measuring the total area enclosed between the function's graph and the x-axis, within those limits.

In the context of our exercise, \( \int_{-1}^{3} \sqrt{4-(x-1)^{2}} dx \), the geometric interpretation is straightforward once we recognize that the integrand, \( \sqrt{4-(x-1)^{2}} \), represents a semicircle. By graphing this function, we can see a semicircle above the x-axis, centered at the point (1, 0) with a radius of 2. The integral evaluates the area of this curved shape from x = -1 to x = 3.

This is a very tangible approach to understanding integrals. Instead of delving into the mechanics of calculus, we can often use geometry to find the same result. Particularly for this type of problem, where the integrand represents a geometric shape, students can better grasp both their calculus lessons and reinforce their knowledge of geometry.

Area Calculation Using Integrals

Integrals don't just help us find areas under curves; they allow us to calculate the area of various shapes defined by functions. For standard functions, such as polynomials, exponentials, or trigonometric functions, we typically rely on the process of integration to find the exact area. However, when the function represents a well-known shape, such as a circle or a semicircle, we can use known geometric formulas to determine the area—an approach exemplified in the exercise.

The area of a circle is a foundational example, which is given by the formula \( A_{circle} = \pi r^2 \). For a semicircle, like the one given in the exercise, we simply take half of this area. When the function represents part of a circle (e.g., a sector), we can also use geometry to aid our calculation.

The technique shown in the solution involves subtracting the area of a sector from the area of a semicircle, both of which have clear geometric formulas, to determine the definitive integral's result. This is a perfect illustration of how integration can be utilized for practical area calculations.

Semi-circle Area in Integration

In our specific exercise, the problem simplifies into calculating the area of a semicircle minus that of a sector. The semicircle is represented by the function \( y = \sqrt{4 - (x-1)^2} \), which upon integration, would yield the area directly underneath it. The area of the full circle would be \( 4\pi \), but since it's a semicircle, we take half, giving us \( 2\pi \).

However, as we're interested in a segment of this semicircle (specifically that between x = -1 and x = 3), we must adjust our calculation. This segment forms a sector, and in geometry, the area of a sector is proportionate to the angle it subtends at the center of the circle relative to the full circle's area.

By subtracting the sector area from the semicircle area, we are left with the area of the integration region, \( \frac{2}{3}\pi \). Thus, integrals not only allow us to calculate areas under arbitrary function curves but also offer a method to deal with more complex geometric shapes – and all of this without relying on the tedious Riemann sum method traditionally used to define integrals.