Problem 22 - Thomas Calculus

Question

In Exercises \(15-22,\) graph the integrands and use areas to evaluate the integrals. $$ \int_{-1}^{1}\left(1+\sqrt{1-x^{2}}\right) d x $$

Step-by-Step Solution

Answer

\(\int_{-1}^{1}\left(1+\sqrt{1-x^{2}}\right) dx = \frac{\pi}{2} + 2\)

Step 1: Graph the Integrand

The integrand is \( 1 + \sqrt{1-x^2} \). Graphing this function involves first understanding its components. \( \sqrt{1-x^2} \) represents the upper half of a circle with radius 1 centered at the origin (as it describes a semicircle above the x-axis from \( x = -1 \) to \( x = 1 \)). The function \( 1 + \sqrt{1-x^2} \) shifts this semicircle up by 1 unit. Hence, the graph for the integrand is a semicircular arc above the line \( y = 1 \) going from \( x = -1 \) to \( x = 1 \).

Step 2: Calculate the Area Under the Semicircle

The area under the semicircle \( \sqrt{1-x^2} \) from \( x = -1 \) to \( x = 1 \) is half the area of a circle with radius 1. The area of a full circle with radius 1 is \( \pi \), hence the area of the semicircle is \( \frac{\pi}{2} \).

Step 3: Calculate the Area of the Rectangle

Since the function \( 1 + \sqrt{1-x^2} \) is shifted up by 1, it adds a rectangular area under the semicircle. This rectangle is from \( x = -1 \) to \( x = 1 \) with a height of 1 (the additional constant \( 1 \)). The width of the rectangle is \( 2 \) and the height is \( 1 \), thus its area is \( 2 \times 1 = 2 \).

Step 4: Add the Areas to Evaluate the Integral

To evaluate the given integral, we sum the areas of the semicircle and the rectangle. The total area is the sum of the semicircle's area \( \frac{\pi}{2} \) and the rectangle's area \( 2 \). Thus, the integral is equal to \( \frac{\pi}{2} + 2 \).

Key Concepts

Area Under Curve

The concept of the area under a curve is fundamental in calculus and is calculated using definite integrals. In essence, finding the area under a curve allows us to determine the accumulation of infinite points within specified bounds. For a function, particularly continuous and bounded on a given interval, the definite integral computes exactly this area beneath the curve and above the x-axis.

In our exercise,

• We a function that consists of a semicircle and a vertical shift by 1.
• The integral from \(-1 \) to \(1 \) represents this total area under the upper shifted semicircle, \(1 + \sqrt{1-x^2}\).
• Calculating this area involves considering both the geometric shape of the semicircle and the additional rectangle created by the vertical shift.

Thus, integrating a function gives a visual and numerical representation of this space through appropriate summation under the curve.

Graphing Integrands

To solve integrals using areas, it's often beneficial to start by visualizing the integrand. Graphing an integrand involves plotting the function to understand its behavior over a specified interval.

For the function \(1 + \sqrt{1-x^2}\), the integrand consists of:

• The component \(\sqrt{1-x^2}\) creates a semicircle on the first quadrant, implying symmetry about the y-axis.
• The graph is shifted upwards by 1, indicated by the \(+1\), moving the semicircular arc to float above y = 1 over the interval \([-1, 1]\).

Visualizing these components can help interpret how each part of a function contributes to the integral's value, aiding in solving it accurately.

Semicircle

A semicircle is half of a circle, and it can be represented analytically as \(\sqrt{1-x^2}\) when considering a unit circle outline. The boundary stays at the top arc from \([-1,1]\) for the x-values.

In the current problem:

• The \(\sqrt{1-x^2}\) forms the base of our function, representing the curve of the semicircle.
• Knowing the properties of a semicircle—a curved line on top, direct symmetries, and half the area of a full circle—helps compute the integral effectively.
• The area of a semicircle with radius 1 is \(\frac{\pi}{2}\), crucial for calculating areas under certain curves.

Recognizing semicircles in integrands allows for intuitive calculation and appreciation of geometric contributions to integrals.

Applications of Integration

Integration has numerous applications beyond pure mathematics, extending into physics, engineering, and various sciences. It serves as a powerful tool to calculate areas, volumes, and other quantities that accumulate continuously.

In simple terms:

• It helps answer practical questions such as "How much?" when dealing with quantities that cannot be easily measured discretely.
• In the context of this exercise, integration demonstrates spatial understanding, calculating how much total area lies under the curve.
• Such techniques are crucial in optimizing areas, predicting system behaviors, and resolving compound problems in technical fields.

Thus, integrating functions is fundamental in understanding and solving real-world problems across diverse domains.