Problem 18 - Calculus Single Variable

Question

Sketch the integrand of the given definite integral over the interval of integration. Evaluate the integral by calculating the area it represents. $$ \int_{1}^{5}\left(\cos ^{2}(x)+\sin ^{2}(x)\right) d x $$

Step-by-Step Solution

Answer

The integral of \(\cos^2(x) + \sin^2(x)\) from 1 to 5 is 4.

Step 1: Simplify the Integrand

Recall the Pythagorean identity: \(\cos^2(x) + \sin^2(x) = 1\). This means our integrand simplifies to \(1\) over the interval \([1, 5]\). Thus, the integral becomes \(\int_{1}^{5} 1 \, dx\).

Step 2: Set Up the Integral

The simplified integral is \(\int_{1}^{5} 1 \, dx\). Over an interval \([a, b]\), the integral of a constant \(c\) is equal to \(c \times (b-a)\).

Step 3: Calculate the Area

The integral we need to calculate is \(\int_{1}^{5} 1 \, dx\). The area represented by this integral is the rectangle formed by multiplying the constant value (which is 1) by the interval length \([1, 5]\): \(1 \times (5 - 1) = 4\).

Step 4: Interpret the Result

The area under the curve \(y = \cos^2(x) + \sin^2(x)\) from \(x = 1\) to \(x = 5\) is calculated to be 4. This area represents the definite integral's evaluation over the specified interval.

Key Concepts

Pythagorean Identity

The Pythagorean identity is a fundamental concept in trigonometry. It states that for any angle \( x \), the expression \( \cos^2(x) + \sin^2(x) = 1 \) always holds true. This identity is derived from the Pythagorean theorem in geometry, which relates the sides of a right triangle. In this context, \( \cos(x) \) and \( \sin(x) \) are equivalent to the adjacent and opposite sides of a right triangle with a hypotenuse of one, respectively.
By using this identity, we can simplify expressions that involve both cosine and sine, as shown in the exercise. Instead of dealing with the more complex terms \( \cos^2(x) + \sin^2(x) \), we simply replace it with 1, greatly simplifying subsequent calculations.

Integrand Simplification

Simplifying the integrand is a crucial step in solving definite integrals more easily. Often, the initial form of the integrand is complex. In this case, the integrand appears as \( \cos^2(x) + \sin^2(x) \).
The Pythagorean identity allows us to effectively simplify this expression to a much more manageable form: 1. After applying this identity, the integral equation becomes \( \int_{1}^{5} 1 \, dx \). This simplification turns a potentially difficult problem into one of straightforward calculation, which is why identifying and employing identities is such a valuable skill.

Area Calculation

Calculating the area under a curve is the core goal of evaluating a definite integral. Once the integrand is simplified, finding this area becomes a straightforward task. When the integrand is a constant value, such as 1, calculating the integral is equivalent to finding the area of a geometric shape.
In our exercise, the definite integral \( \int_{1}^{5} 1 \, dx \) represents a rectangle. The height of the rectangle is the constant value 1 (from the simplified integrand), and the width is the difference between the upper and lower bounds of integration, 5 and 1, respectively. Therefore, the area, or the value of the definite integral, is calculated as:\[ 1 \times (5 - 1) = 4. \]This rectangle reflects the area under the curve over the specified interval.

Interval of Integration

The interval of integration determines the section of the x-axis over which the area under the curve is to be calculated. In definite integrals, you will often see limits at the bottom and top of the integral sign, like \( \int_{a}^{b} \).
For this particular exercise, the interval is from \( x = 1 \) to \( x = 5 \). These endpoints not only define the range on the x-axis but also serve as boundaries of the geometric shape—here, a rectangle—that we are considering when calculating the area. The length of this interval, calculated as \( b - a = 5 - 1 = 4 \), is multiplied by the constant function value to get the area. Thus, understanding the interval is critical for solving any problem involving definite integrals.