Problem 24 - Calculus Early Transcendentals

Question

Evaluate the following integrals using the Fundamental Theorem of Calculus. Explain why your result is consistent with the figure. $$\int_{-\pi / 4}^{7 \pi / 4}(\sin x+\cos x) d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral of the function \((\sin x + \cos x)\) over the interval \([-\pi/4, 7\pi/4]\). Answer: The definite integral of the function \((\sin x + \cos x)\) over the interval \([-\pi/4, 7\pi/4]\) is equal to \(2\sqrt{2}\).

Step 1: Find the antiderivative of the given function

We will find the antiderivative of \((\sin x + \cos x)\): Let \(F(x)\) be the antiderivative of \((\sin x + \cos x)\). After differentiating the function with respect to \(x\), we get: $$F'(x) = \sin x + \cos x$$ Now, we need to find the function \(F(x)\). To do so, we integrate \(F'(x)\) with respect to \(x\). In this case, we can integrate term by term: $$F(x) = \int (\sin x + \cos x) dx = \int \sin x dx + \int \cos x dx$$ Now, we find the antiderivatives of each term: $$F(x) = -\cos x + \sin x + C$$ Where \(C\) is the constant of integration.

Step 2: Evaluate the definite integral using the Fundamental Theorem of Calculus

Now that we have the antiderivative, we can apply the Fundamental Theorem of Calculus to evaluate the definite integral: $$\int_{-\pi/4}^{7\pi/4} (\sin x + \cos x) dx = F(7\pi/4) - F(-\pi/4)$$ Now, we substitute the values into \(F(x)\) and calculate the result: $$F(7\pi/4) = -\cos(7\pi/4) + \sin(7\pi/4) = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}$$ $$F(-\pi/4) = -\cos(-\pi/4) + \sin(-\pi/4) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}$$ Now, we subtract the two values: $$F(7\pi/4) - F(-\pi/4) = \left[-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right] - \left[-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right]$$ After simplifying the expression, we obtain: $$F(7\pi/4) - F(-\pi/4) = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$$ Thus, the definite integral is: $$\int_{-\pi/4}^{7\pi/4} (\sin x + \cos x) dx = 2\sqrt{2}$$

Key Concepts

Definite Integral

The concept of a definite integral is integral (pun intended!) to understanding calculus. It represents the 'net area' under a curve between two points on the x-axis. In other words, if you have a function and you want to calculate the total accumulation of that function's value between two bounds, you would use a definite integral.

For example, say we want to measure the total change in distance for a moving object over a period of time given its speed at any point in time. This is where the definite integral shines. By integrating the speed function over a certain time interval, we'd get the total distance traveled.

In the exercise you're working with, the definite integral is used to combine the accumulated effects of both sine and cosine between \( -\frac{\pi}{4} \) and \( 7\frac{\pi}{4} \). Since the sine and cosine functions are periodic and symmetric, the integral will take into account their oscillations over the interval. The result, \( 2\sqrt{2} \), represents the algebraic sum of all these oscillations over the given period.

Antiderivative

Peering into the world of calculus, the antiderivative–also known as the indefinite integral–is essentially the reverse of differentiation. When you find the derivative of a function, you're calculating the rate at which something changes. The antiderivative, on the other hand, takes you back from the rate of change to the original quantity.

If we relate this to a real-life scenario: if you're given the rate at which a balloon loses air over time, finding the antiderivative would give you the function that represents the total air volume in the balloon at any given time, before it began to deflate.

In our specific exercise, finding the antiderivative of \( \sin x + \cos x \) resulted in \( -\cos x + \sin x + C \), which played a critical role in applying the Fundamental Theorem of Calculus. This theorem marries the derivative and the integral, stating that if you take the antiderivative of a function and then evaluate it at the upper and lower bounds, you've computed the definite integral.

Sine and Cosine Functions

Let's dip our toes into the refreshing pool of trigonometric functions, focusing on the sine and cosine waves. These functions are perhaps most familiar as the mathematical description of oscillations, waves, and circular motion.

We encounter them almost everywhere: the periodic rise and fall of sea waves, the oscillation of a pendulum, even the fluctuations of alternating current in electronics are all described by these fundamental functions.

Under the hood, the sine function, denoted as \( \sin x \), measures the vertical component of a point rotating on a unit circle, while the cosine function, \( \cos x \), measures the horizontal component. They are harmonious in their pattern, with one being the phase-shifted version of the other.

In the fundamental theorem of calculus exercise, the integrals of sine and cosine functions become crucial concepts. Since the interval from \( -\frac{\pi}{4} \) to \( 7\frac{\pi}{4} \) encompasses more than a full cycle of the functions, the integral includes areas above and below the x-axis, which in this scenario, happen to cancel out symmetrically, leaving behind the neatly packaged result of \( 2\sqrt{2} \).