Question

Sketch the region whose area is represented by the definite integral. Then use a geometric formula to evaluate the integral. $$ \int_{0}^{2} \sqrt{4-x^{2}} d x $$

Short answer

The area represented by the integral is the quarter of a circle of radius 2, the area is then \( A = \pi \).

Step-by-step solution

Sketch the region

Begin with sketching the region whose area is represented by this definite integral. The equation under integration, \( \sqrt{4-x^{2}} \), is the equation for the upper half of a circle with radius \( r = 2 \) centered at the origin. The integral from 0 to 2 indicates this represents a quarter of the circle in the positive quadrant of a graph.

Recognize the geometric shape

By looking at the figure, you can see that the region that the function sketches from \( x = 0 \) to \( x = 2 \) is a quarter of a circle with radius \( r = 2 \).

Apply geometric formula

Apply the formula for area of a circle which is \( A = \pi \ r^{2} \). But since we only have a quarter of that circle, we need to divide the formula by 4, thus we use \( A = (1/4) \ \pi \ r^{2} = (1/4) \ \pi \ (2)^{2} = \pi \).

Verify result

Substitute the limits of integration \(x = 0\) and \(x = 2\) into the calculated area A. Verify that the area under the curve for these limits equals \(A = \pi\). This confirms that the geometrically computed area is same as the definite integral's value.

Key concepts

Geometric Interpretation of Definite Integrals

When it comes to understanding definite integrals, it can be really beneficial to think about them in terms of geometry. A definite integral essentially provides the total 'accumulation' of a quantity, and where calculus meets geometry, this often translates to the area under a curve on a graph between two points.

Take the exercise above, for instance, which asks you to evaluate the integral of the function \( \(\sqrt{4-x^{2}}\) \) from \( x = 0 \) to \( x = 2 \). This function describes a semicircle when plotted, with a radius defined by the square root. By recognising this shape, we can cleverly deduce the area described by the function without performing complex integration: since the area under the curve from 0 to 2 outlines a quarter of a circle, the integral that seems algebraically challenging at first glance, simplifies to the well-known area formula of a circle. It's a clear example of how visualising integrals can simplify computations considerably, making them more approachable and less frightening.

Area Formulas in Calculus

Area formulas in calculus are your best friends when faced with a definite integral that has a geometric interpretation. Knowing these formulas enables you to evaluate integrals quickly and accurately, especially when the integral represents a familiar shape like a rectangle, triangle, circle, or trapezoid.

In the context of our example, the area represented by the definite integral is for a quarter circle. The area of a full circle is given by \( A = \pi r^{2} \) where \( r \) is the radius. Since we're only dealing with a quarter of a circle, we divide the total area by 4, leading to \( A = \frac{1}{4} \pi r^{2} \) for our quarter circle. Substituting \( r = 2 \), we get \( A = \pi \), which coincides with the evaluation of the definite integral given in the exercise. Knowing these formulas can turn what could be an intimidating calculus problem into a simple geometry one.

Integration in Algebraic and Geometric Contexts

Integration is an operation in calculus that's often used to combine small data points into a whole, and it's useful in both algebraic and geometric contexts. Algebraically, an integral is the sum of infinitesimal parts that add up to a function's area of influence between two points, while geometrically, an integral can sometimes represent a tangible shape and area we can calculate using geometry.

In the case of our exercise, the function \( \sqrt{4-x^{2}} \) describes a curve that, when drawn, creates a boundary for a geometric shape. The act of integrating this function between 0 and 2 is algebraically complex, but by connecting it to a geometric context—recognizing that this boundary forms a quarter of a circle—we shift the task from integration to applying the area formula of a circle. This duality that integration possesses enriches the study of calculus because it allows for creative problem-solving where both algebra and geometry are interwoven to achieve a solution.