Question

Approximate the value of the given definite integral by using the first 4 nonzero terms of the integrand's Taylor series. $$\int_{0}^{\pi^{2} / 4} \cos (\sqrt{x}) d x$$

Short answer

The approximate value is \(\frac{\pi^2}{4} - \frac{\pi^4}{128} + \frac{\pi^6}{4608} - \frac{\pi^8}{737280}\).

Step-by-step solution

Find the Taylor Series of the Integrand

The function we are integrating is \(\cos(\sqrt{x})\). First, we need the Taylor series expansion of \(\cos(x)\) around \(x = 0\), which is: \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \] Substitute \(x = \sqrt{x}\) to get the series for \(\cos(\sqrt{x})\): \[ \cos(\sqrt{x}) = 1 - \frac{x}{2} + \frac{x^2}{4!} - \cdots \]

Approximate the Integrand Using the First 4 Terms

We need to stop at the first four non-zero terms. The Taylor series for \(\cos(\sqrt{x})\) up to four non-zero terms is: \[ 1 - \frac{x}{2} + \frac{x^2}{24} - \frac{x^3}{720} \] We will use this approximated integrand to perform the integration.

Integrate the Approximated Series Term by Term

Now we integrate each term separately over the range \(0\) to \(\frac{\pi^2}{4}\):- \(\int_{0}^{\pi^2/4} 1 \, dx = x \bigg|_0^{\pi^2/4} = \frac{\pi^2}{4} \)- \(\int_{0}^{\pi^2/4} \frac{x}{2} \, dx = \frac{1}{2} \cdot \frac{x^2}{2} \bigg|_0^{\pi^2/4} = \frac{1}{8} \left(\frac{\pi^4}{16}\right) = \frac{\pi^4}{128} \)- \(\int_{0}^{\pi^2/4} \frac{x^2}{24} \, dx = \frac{1}{24} \cdot \frac{x^3}{3} \bigg|_0^{\pi^2/4} = \frac{1}{72} \left(\frac{\pi^6}{64}\right) = \frac{\pi^6}{4608} \)- \(\int_{0}^{\pi^2/4} \frac{x^3}{720} \, dx = \frac{1}{720} \cdot \frac{x^4}{4} \bigg|_0^{\pi^2/4} = \frac{1}{2880} \left(\frac{\pi^8}{256}\right) = \frac{\pi^8}{737280} \)

Combine the Results of Integration

Add the results from step 3 to get the approximation of the integral:\[ \frac{\pi^2}{4} - \frac{\pi^4}{128} + \frac{\pi^6}{4608} - \frac{\pi^8}{737280} \].

Key concepts

Definite Integral

A definite integral is used to find the area under a curve within a certain interval. When we write \( \int_{a}^{b} f(x) \, dx \), we are computing the area under the curve of \( f(x) \) from \( x = a \) to \( x = b \). In the exercise, we have used a definite integral: \( \int_{0}^{\pi^{2}/4} \cos(\sqrt{x}) \, dx \). This expression signifies that we want to determine the net area between the graph of \( \cos(\sqrt{x}) \) and the x-axis from \( x = 0 \) to \( x = \pi^{2}/4 \).

• Upper and Lower Bounds: Definite integrals require limits of integration. Here, the lower bound is 0 and the upper bound is \( \pi^{2}/4 \).
• Function Area: The integral provides a measure of the total area between the function and the x-axis for the given interval.
• Outcome: The result can be positive or negative, indicating the direction of the area with respect to the x-axis.

Definite integrals are foundational in calculus, allowing us to compute areas and solve various applications in real-world contexts.

Approximation

Approximation is a method of finding an approximate value when exact calculations are difficult or impossible. In calculus, the Taylor series is a powerful tool for approximating functions using a sum of polynomial terms. This approximation stems from using derivatives to represent and simplify complex functions.
In the given exercise, instead of evaluating the function \( \cos(\sqrt{x}) \) directly, we use a Taylor series expansion to find a simpler form that is easier to integrate. The Taylor series is: \[ \cos(\sqrt{x}) \approx 1 - \frac{x}{2} + \frac{x^2}{24} - \frac{x^3}{720} \]

• Taylor Series: A representation of a function as an infinite sum of terms, calculated from the values of its derivatives at a single point.
• Truncating the Series: We only use the first four non-zero terms of the series to make the problem manageable.
• Purpose: By approximating \( \cos(\sqrt{x}) \), we reduce complexity, making the integration plausible.

Approximation is crucial in calculus, offering efficient solutions when dealing with complex functions or integrals.

Cosine Function

The cosine function, written as \( \cos(x) \), is a fundamental trigonometric function that describes the relationship between the sides of a right-angled triangle. Cosine is periodic, symmetric, and its range is between -1 to 1. Here, it plays a pivotal role in our integral calculus exercise.
For \( \cos(\sqrt{x}) \), we are evaluating the cosine of the square root of \( x \), which adds an interesting twist to our problem. We use the Taylor series of \( \cos(x) \) to transform it into something integrable. The Taylor series of \( \cos(x) \) around \( x=0 \) is given by: \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \]

• Symmetry: Cosine is an even function, which means \( \cos(-x) = \cos(x) \).
• Use in Integration: Its periodic and smooth nature makes it suitable for approximations.

The cosine function's predictable pattern and properties make it ideal for expressing more complex relationships through approximations.

Integration by Parts

Integration by parts is a powerful technique that comes from the product rule for differentiation. It's used to transform the integral of a product of functions into simpler forms, making them easier to evaluate or approximate. While not directly used in this exercise, understanding it provides valuable insights into derivative and integral manipulation.
The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]It requires choosing parts of the integral as \( u \) and \( dv \), differentiating \( u \) to get \( du \), and integrating \( dv \) to find \( v \). What follows is the application of this formula to simplify and solve integrals in a stepwise manner.

• Product of Functions: Ideal when dealing with integrals involving products of polynomials, exponentials, or trigonometric functions.
• Simplification: Converts challenging integrals into more manageable ones.

Although not needed for this particular exercise, integration by parts complements Taylor series to solve more complex integrals when approximation is not sufficient.