Problem 12 - Thomas Calculus

Question

In Exercises \(1-16,\) find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. $$\text { a. }\pi \cos \pi x \quad \text { b. } \frac{\pi}{2} \cos \frac{\pi x}{2} \quad \text { c. } \cos \frac{\pi x}{2}+\pi \cos x$$

Step-by-Step Solution

Answer

a: \(\sin(\pi x)+C\), b: \(\sin\left(\frac{\pi x}{2}\right)+C\), c: \(\frac{2}{\pi}\sin\left(\frac{\pi x}{2}\right)+\sin(x)+C\).\)

Step 1: Consider Part a

To find the antiderivative of the function \( f(x) = \pi \cos(\pi x) \), remember:\[ \int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C \]. Here, \( a = \pi \). Therefore, the antiderivative is:\[ F(x) = \frac{1}{\pi} \sin(\pi x) + C \]. Since the function is multiplied by \( \pi \), our antiderivative becomes:\[ F(x) = \sin(\pi x) + C \].

Step 2: Verify Part a by Differentiation

Differentiate the function \( F(x) = \sin(\pi x) \):\[ F'(x) = \pi \cos(\pi x) \]. This matches the given function \( \pi \cos(\pi x) \), confirming our antiderivative is correct.

Step 3: Consider Part b

Find the antiderivative of \( f(x) = \frac{\pi}{2} \cos \left( \frac{\pi x}{2} \right) \). Using \[ \int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C \] with \( a = \frac{\pi}{2} \), we find:\[ F(x) = \frac{1}{\left(\frac{\pi}{2}\right)} \sin \left( \frac{\pi x}{2} \right) + C = \frac{2}{\pi} \sin \left( \frac{\pi x}{2} \right) + C \]. Since \( \frac{\pi}{2} \) is multiplied by this result, the antiderivative becomes \( \sin \left( \frac{\pi x}{2} \right) + C \).

Step 4: Verify Part b by Differentiation

Differentiate \( F(x) = \sin \left( \frac{\pi x}{2} \right) \):\[ F'(x) = \frac{\pi}{2} \cos \left( \frac{\pi x}{2} \right) \]. This matches the original function \( \frac{\pi}{2} \cos \left( \frac{\pi x}{2} \right) \), confirming the correctness of the antiderivative.

Step 5: Consider Part c

The function is \( f(x) = \cos \left( \frac{\pi x}{2} \right) + \pi \cos(x) \). Antiderivatives for \( \cos \left( \frac{\pi x}{2} \right) \) and \( \pi \cos(x) \) are \( \frac{2}{\pi} \sin \left( \frac{\pi x}{2} \right) + C_1 \) and \( \sin(x) + C_2 \) respectively. The combined antiderivative is:\[ F(x) = \frac{2}{\pi} \sin \left( \frac{\pi x}{2} \right) + \sin(x) + C \].

Step 6: Verify Part c by Differentiation

Differentiate \( F(x) = \frac{2}{\pi} \sin \left( \frac{\pi x}{2} \right) + \sin(x) \):\[ F'(x) = \cos \left( \frac{\pi x}{2} \right) + \pi \cos(x) \]. This matches the original function, confirming our antiderivative: \( \frac{2}{\pi} \sin \left( \frac{\pi x}{2} \right) + \sin(x) \).

Key Concepts

Cosine Function

The cosine function, often symbolized as \( \cos(x) \), is a fundamental trigonometric function. It is periodic, meaning it repeats its values in regular intervals, specifically every \( 2\pi \) radians. The cosine function starts at 1 when \( x = 0 \) and then oscillates between -1 and 1.

• Cosine is an even function, so \( \cos(-x) = \cos(x) \).
• The graph of the cosine function is a smooth wave-like pattern, which is why it's used in modeling wave phenomena.

The standard form is modified by coefficients to stretch or compress the wave. For example, in the function \( \pi \cos(\pi x) \), the coefficient \( \pi \) changes both the frequency and amplitude of the wave.Understanding these properties is crucial as they directly affect how we find the antiderivative or differentiate the function.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative measures how a function changes when its input changes.

• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• The process of differentiation follows specific rules, such as the product rule, quotient rule, and chain rule.

In the context of the exercise, after finding the antiderivative of each function, differentiation is used to verify the solution. For instance, by differentiating \( F(x) = \sin(\pi x) \), we should return to the original function \( \pi \cos(\pi x) \). This verification step is crucial, showing the relationship between a function and its derivative is consistent. Differentiation helps check if the correct antiderivative (integral) was determined.

Integral Calculus

Integral calculus focuses on the concept of integrals, which can be thought of as the reverse process to differentiation. An antiderivative is a specific type of integral, that helps find the area under the curve of a function.

• Finding an antiderivative involves determining a function whose derivative is the given function.
• The integral of \( \cos(ax) \) is \( \frac{1}{a} \sin(ax) + C \), where \( C \) is the constant of integration.

In exercises like ours, we find the antiderivative of functions such as \( \pi \cos(\pi x) \). We use integral calculus techniques to show that \[ \int \pi \cos(\pi x) \, dx = \sin(\pi x) + C \].This process demonstrates how integral calculus allows us to recover the original function from its rate of change, highlighting the intrinsic connection between differentiation and integration.