Problem 34 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral and interpret it as a difference of areas. Illustrate with a sketch. \(\int_{-\pi / 2}^{2 \pi} \cos x d x\)

Step-by-Step Solution

Answer

The integral evaluates to 1, representing the net area under the curve from \\\( -\\pi/2 \\\\) to \\\( 2\\pi \\\\). It's the net area of one full cycle of \\( \cos x \\\) with some cancellation of negative areas.

Step 1: Identify the function and interval

We need to evaluate the integral of the function \( \cos x \) over the interval \( [-\pi/2, 2\pi] \). This means finding the area under the curve of \( \cos x \) from \( x = -\pi/2 \) to \( x = 2\pi \).

Step 2: Compute the indefinite integral

The antiderivative of \( \cos x \) is \( \sin x \). This will help us evaluate the definite integral. Thus, we have:\[\int \cos x \, dx = \sin x + C\]

Step 3: Evaluate the definite integral

Now, apply the limits to the antiderivative:\[\int_{-\pi / 2}^{2 \pi} \cos x \ \, dx = \left[ \sin x \right]_{-\pi / 2}^{2 \pi}\]Substituting the limits into the antiderivative, we get:\[\sin(2\pi) - \sin(-\pi/2) = 0 - (-1) = 1\]

Step 4: Interpret the integral result as an area under the curve

The integral evaluates to 1, which represents the net area under the curve of \( \cos x \) from \( x = -\pi/2 \) to \( x = 2\pi \). This encompasses one full cycle of the cosine function, which includes positive area from \( 0 \) to \( 2\pi \) and negative area from \( -\pi/2 \) to \( 0 \). The final result is the net area, subtracting the absolute values of negative contributions from the positive areas.

Step 5: Illustrate the integral with a sketch

Sketch a graph of \( \cos x \) from \( x = -\pi/2 \) to \( x = 2\pi \). The graph will show the curve oscillating between -1 and 1, with one full cycle from \( 0 \) to \( 2\pi \) and a partial cycle from \( -\pi/2 \) to \( 0 \), producing a net positive area of 1. The shaded area under the curve from this range will help visualize the accumulated area.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in integral calculus used to calculate the accumulation of quantities. They provide the means to compute areas under a function's curve within a specific interval. The process involves evaluating the integral through an antiderivative, which is then applied to the given limits of the interval.
For instance, in the original exercise, the definite integral \( \int_{-\pi / 2}^{2 \pi} \cos x \, dx \) is evaluated over the interval \([-\pi/2, 2\pi]\). This calculates the net area beneath the cosine curve within these bounds.

• The antiderivative, \( \sin x + C \), is used to determine areas when limits are applied.
• The definite integral eliminates the constant \( C \) by using the fundamental theorem of calculus.

Ultimately, definite integrals are a powerful tool for quantifying changes, capturing the total effect of a function across a defined segment.

Cosine Function

The cosine function \( \cos x \) is one of the basic trigonometric functions that depicts periodic oscillations. It manifests as a wave that oscillates between -1 and 1, repeating every \( 2\pi \) radians.
Attributes of the cosine function include:

• Periodicity: It has a period of \( 2\pi \), meaning it repeats its pattern every \( 2\pi \) units along the x-axis.
• Amplitude: The maximum and minimum values of \( \cos x \) are 1 and -1, defining its amplitude.
• Even Function: Cosine is an even function, hence \( \cos(-x) = \cos(x) \).

Understanding these characteristics is crucial when evaluating integrals involving \( \cos x \), as evidenced in the exercise. Observing its graph aids in visualizing where the curve dips below or rises above the x-axis, critical for measuring positive and negative areas.

Net Area Under Curve

The net area under a curve refers to the total area that is calculated when accounting for both positive and negative regions under a function. This concept is crucial in definite integrals, which tally the areas above and below the x-axis.
In the case of the original exercise, the integral of \( \cos x \) over \([-\pi/2, 2\pi]\) demonstrates the idea of net area. Here’s how it works:

• From \( -\pi/2 \) to 0: the curve is beneath the x-axis, contributing negative area.
• From 0 to \( 2\pi \): it is above the x-axis, adding positive area.

The final outcome of 1 represents the net area, meaning the negative area is subtracted from the total positive area accumulated from \( -\pi/2 \) to \( 2\pi \). This concept efficiently reconciles varying contributions of the curve to afford a clearer representation of the overall impact.