Problem 51 - Calculus

Question

Calculate \(\int_{1}^{1+\pi}|\cos x| d x.\)

Step-by-Step Solution

Answer

The integral of \(|\cos x|\) from \(x = 1\) to \(x = 1 + \pi\) is 0.

Step 1: Analyze the Limits of Integration

The problem asks us to find the definite integral of \(|\cos x|\) from \(x = 1\) to \(x = 1 + \pi\). Given that \(1 + \pi\) is slightly more than two full cycles of \(\pi\) units for cosine, we need to break this into known intervals of cosine behavior (positive or negative) for \([1, 1+\pi]\).

Step 2: Plot and Identify Critical Points

On the interval \([1, 1+\pi]\), determine where \(\cos x\) changes sign. Since \(\cos x\) becomes negative between \(\frac{\pi}{2}\) (approx. 1.57) to \(\frac{3\pi}{2}\) (approx. 4.71), calculate specific values: \(\cos 1 > 0\) and \(\cos (1+\pi) < 0\). Critical points: \(x = 1\), \(x = \frac{3\pi}{2}\).

Step 3: Split the Integral at Critical Points

Split the integral into segments: 1) \([1, \pi/2]\), and 2) \([\pi/2, 1+\pi]\). On \([1, \pi/2]\), \(\cos x\) is positive, thus \(|\cos x| = \cos x\). On \([\pi/2, 1+\pi]\), \(\cos x\) is negative, thus \(|\cos x| = -\cos x\).

Step 4: Calculate the Integral for \([1, \pi/2]\)

Evaluate the integral \(\int_{1}^{\pi/2} \cos x \, dx\). Using the identity \(\int \cos x \, dx = \sin x\), we find: \[\text{Part 1: } \int_{1}^{\pi/2} \cos x \, dx = \sin x \Big|_{1}^{\pi/2} = \sin(\pi/2) - \sin(1) = 1 - \sin(1).\]

Step 5: Calculate the Integral for \([\pi/2, 1+\pi]\)

Evaluate the integral \(\int_{\pi/2}^{1+\pi} -\cos x \, dx\), which is equal to \(-\int_{\pi/2}^{1+\pi} \cos x \, dx\). Again, use the identity \(\int \cos x \, dx = \sin x\): \[\text{Part 2: } -\int_{\pi/2}^{1+\pi} \cos x \, dx = -(\sin(1+\pi) - \sin(\pi/2)) = -(\sin(1+\pi) - 1).\] Simplifying: when \(x = 1 + \pi\), \(\sin(1+\pi) = -\sin(1)\) due to periodicity of sine. Thus, Part 2 becomes: \[\sin(1) - 1.\]

Step 6: Combine Results for Final Solution

Combine the two parts to find the solution to \(\int_{1}^{1+\pi}|\cos x| \, dx\) by adding results from step 4 and step 5: \[\text{Result} = (1 - \sin(1)) + (\sin(1) - 1) = 0.\] This shows that the integral of \(|\cos x|\) over this interval results in a net zero change, which aligns with full cycles of cosine.

Key Concepts

Absolute Value Functions

Absolute value functions transform or translate input values into their non-negative forms. This means the distance of each value from zero is calculated without considering whether it is positive or negative. In our specific integral, \[\int_{1}^{1+\pi} |\cos x| \, dx,\]we are working with the absolute value of the cosine function. Cosine, being a periodic function, oscillates between -1 and 1. Therefore,

• Where cosine is positive, the absolute value doesn't change it.
• Where cosine is negative, the absolute value converts the negative portion of cosine into positive, essentially mirroring it above the x-axis.

This property of absolute values is crucial when evaluating definite integrals. It demands determination of segments where the function inside the absolute value is positive versus negative. In this case, we had certain intervals where the cosine value naturally changed its sign (from positive to negative), which led to the necessity of breaking up the integral and recalculating its value in pieces, rather than trying to integrate it as a whole at once.

Trigonometric Functions

Trigonometric functions like cosine and sine play an indispensable role in calculus, especially in problems involving periodic behavior. Let’s delve into some properties that are particularly relevant:

• Periodicity: The cosine function repeats itself every full cycle of \(2\pi\), returning to its original value. However, on smaller intervals such as \([1, 1+\pi]\), understanding where cosine becomes negative or positive is key to solving integrals involving absolute values.
• Sign Changes: Within the stated limits, understanding where \(\cos x\) crosses from positive to negative is necessary. This happens near special angles like \(\pi/2\) and \(3\pi/2\), where the cosine precisely changes sign, leading to mirrored parts in its graph.

Graphs can be immensely helpful when dealing with trigonometric functions. They assist in visualizing how the function behaves over specific intervals, highlighting at what points changes occur. In calculus, such visual aids provide insight into how shifts, peaks, or dips in the function may affect integral calculations, especially when dealing with absolute values.

Integration by Parts

Integration by parts is a useful technique in calculus when dealing with products of functions. Although not directly applied in the earlier solution, it's beneficial to understand as it showcases methods where direct integration is tough. Integration by Parts is derived from the product rule for differentiation and can be stated as:\[\int u \, dv = uv - \int v \, du,\]where \(u\) and \(dv\) are parts of the original function to integrate.
This formula is particularly useful when:

• The integrand is a product of two distinct functions, like polynomials and exponentials, or polynomials and trigonometric functions.
• One of the functions, when differentiated, simplifies the integral.

Understanding when to apply this technique can dramatically simplify complex integrals. Although this wasn't necessary for handling the given problem – due to the simpler nature of the function \(|\cos x|\) without direct multiplication by another distinct function – it's a critical component in the broader toolkit of methods available for definite integration.