Question

Find the area of the region enclosed by the curve \(y=x \cos x\) and the \(x\) -axis for \(\frac{11 \pi}{2} \leq x \leq \frac{13 \pi}{2} .\) (Express the answer in exact form.)

Short answer

Evaluate \(\int_{rac{23\pi}{4}}^{\frac{25\pi}{4}} -x \cos x \, dx\) using integration by parts and compute the definite values to find the area.

Step-by-step solution

Identify the Intersection Points

Since we need the area between the curve \(y = x \cos x\) and the \(x\)-axis, we first find the points where the curve intersects the \(x\)-axis. This occurs when \(y = 0\), so we solve \(x \cos x = 0\). This equation is true when either \(x = 0\) or \(\cos x = 0\). For the given range \(\frac{11\pi}{2} \leq x \leq \frac{13\pi}{2}\), \(\cos x = 0\) at \(x = \frac{23\pi}{4}\) and \(x = \frac{25\pi}{4}\).

Set Up the Integral

The region enclosed by the curve and the \(x\)-axis is determined by the integral of the absolute value of \(y = x \cos x\) over the intervals defined by the intersection points. The integral we need to compute is:\[\text{Area} = \int_{\frac{23\pi}{4}}^{\frac{25\pi}{4}} |x \cos x| \, dx\]

Split the Integral if Needed

Because \(\cos x\) switches sign at odd multiples of \(\frac{\pi}{2}\), we must consider the sign of the function over the interval. For our specific part from \(\frac{23\pi}{4}\) to \(\frac{25\pi}{4}\), there is no sign change within each interval of \(\pi/2\) since the range considers values beyond two full intervals of the cosine function oscillation in this range.

Evaluate the Integral

For the entire problem, evaluate the definite integral considering the absolute value over the identified range:\[|x \cos x| = \begin{cases} -x \cos x, & \text{if } \frac{23\pi}{4} \leq x \leq \frac{25\pi}{4}\end{cases}\]The integral becomes:\[\int_{\frac{23\pi}{4}}^{\frac{25\pi}{4}} -x \cos x \, dx\]

Find the Anti-derivative

To solve \(\int -x \cos x \, dx\), we use integration by parts with \(u = x\), \(dv = -\cos x \, dx\), which gives \(du = dx\) and \(v = -\sin x\). The integration by parts formula is \(\int u \, dv = uv - \int v \, du\).\[\int -x \cos x \, dx = -x \sin x + \int \sin x \, dx\]\[= -x \sin x + \cos x\]

Calculate the Definite Integral

Use the antiderivative found in step 5 to evaluate the area:\[\left[ -x \sin x + \cos x \right]_{\frac{23\pi}{4}}^{\frac{25\pi}{4}} \]Evaluate at the bounds:\[(-x \sin x + \cos x) \bigg|_{\frac{25\pi}{4}} - (-x \sin x + \cos x) \bigg|_{\frac{23\pi}{4}}\]

Simplify and Obtain the Exact Area

After computing the expression from the definite integral, ensure to simplify each term by calculating the sine and cosine values at \(\frac{23\pi}{4}\) and \(\frac{25\pi}{4}\), considering their positions in the unit circle. Obtain the final expression for the area.

Key concepts

Integration by Parts

Integration by parts is a useful technique in calculus, especially when dealing with products of functions, such as polynomials and trigonometric functions. The formula is based on the product rule for differentiation and is stated as:\[\int u\, dv = uv - \int v\, du\]To apply integration by parts, you follow these steps:

• Select which part of the integral should be \(u\) and which should be \(dv\).
• Differentiate \(u\) to find \(du\).
• Integrate \(dv\) to get \(v\).
• Substitute into the formula: \(uv - \int v\, du\).

In our example, we solved the integral \(\int -x \cos x\, dx\) by choosing \(u = x\) and \(dv = -\cos x\ dx\). This led to \(du = dx\) and \(v = -\sin x\), and applying the formula allows us to find the antiderivative. It simplifies the process of evaluating integrals involving products significantly.

Definite Integrals

Definite integrals are fundamental in finding the exact area under a curve between two limits. These integrals not only give you the area but also take into account the concept of signed areas when dealing with functions that cross the x-axis. A definite integral is represented as:\[\int_{a}^{b} f(x) \, dx\]Here, \(a\) and \(b\) are the limits of integration, and \(f(x)\) is the function under consideration.To compute a definite integral:

• Find the indefinite integral (antiderivative) of the function.
• Evaluate this antiderivative at the upper limit \(b\) and lower limit \(a\).
• Subtract the results: \(F(b) - F(a)\).

In the problem, we evaluated the definite integral of \(|x \cos x|\) by applying these steps to arrive at the solution. We carefully considered the sign of the function due to the absolute value, making sure to use the correct integral expression.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are periodic and oscillating functions essential in many calculus problems. These functions have distinct properties and are defined on the unit circle. Key properties of trigonometric functions include:

• The period of \(\cos(x)\) and \(\sin(x)\) is \(2\pi\), meaning they repeat every \(2\pi\).
• \(\cos(x)\) is the x-coordinate and \(\sin(x)\) is the y-coordinate of a point on the unit circle.
• They intersect the x-axis at specific angles (\(\pi/2\), \(3\pi/2\), etc. for \(\cos(x)\)).

In the exercise, \(\cos(x)\) changed signs at specific points, which affected the integral calculation. Understanding these properties helps in correctly setting up and evaluating integrals involving trigonometric functions. Recognizing the behavior of sine and cosine at different values is crucial for precise computation.