Question

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

Short answer

Approximate the integral: \( \frac{\pi^{3/2}}{3} - \frac{\pi^{7/2}}{42} + \frac{\pi^{11/2}}{1320} - \frac{\pi^{15/2}}{75600} \).

Step-by-step solution

Find the Taylor Series of the Function

The first task is to find the Taylor series of the function \( \sin(x^2) \). The Taylor series for \( \sin(x) \) about 0 is \( \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \). To find \( \sin(x^2) \), we substitute \( x^2 \) instead of \( x \), giving us \( \sum_{n=0}^{\infty} (-1)^n \frac{(x^2)^{2n+1}}{(2n+1)!} \).

Determine the First Four Nonzero Terms

We need to expand the series and determine the first four nonzero terms. Calculating these we have:1. First term: \( x^2 \)2. Second term: \( -\frac{x^6}{3!} = -\frac{x^6}{6} \)3. Third term: \( \frac{x^{10}}{5!} = \frac{x^{10}}{120} \)4. Fourth term: \( -\frac{x^{14}}{7!} = -\frac{x^{14}}{5040} \).Thus, the approximation for \( \sin(x^2) \) is \( x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \frac{x^{14}}{5040} \).

Integrate the Taylor Series Terms

Integrate term by term over the interval from 0 to \( \sqrt{\pi} \):1. \( \int_{0}^{\sqrt{\pi}} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{\sqrt{\pi}} \)2. \( \int_{0}^{\sqrt{\pi}} -\frac{x^6}{6} \, dx = -\frac{1}{6} \left[ \frac{x^7}{7} \right]_{0}^{\sqrt{\pi}} \)3. \( \int_{0}^{\sqrt{\pi}} \frac{x^{10}}{120} \, dx = \frac{1}{120} \left[ \frac{x^{11}}{11} \right]_{0}^{\sqrt{\pi}} \)4. \( \int_{0}^{\sqrt{\pi}} -\frac{x^{14}}{5040} \, dx = -\frac{1}{5040} \left[ \frac{x^{15}}{15} \right]_{0}^{\sqrt{\pi}} \).

Evaluate the Integrals

Now, compute the values:1. \( \frac{(\sqrt{\pi})^3}{3} = \frac{\pi^{3/2}}{3} \)2. \( -\frac{1}{6} \times \frac{(\sqrt{\pi})^7}{7} = -\frac{\pi^{7/2}}{42} \)3. \( \frac{1}{120} \times \frac{(\sqrt{\pi})^{11}}{11} = \frac{\pi^{11/2}}{1320} \)4. \( -\frac{1}{5040} \times \frac{(\sqrt{\pi})^{15}}{15} = -\frac{\pi^{15/2}}{75600} \).

Approximate the Integral

Add the evaluated terms together:\[ \int_{0}^{\sqrt{\pi}} \sin(x^2) \approx \frac{\pi^{3/2}}{3} - \frac{\pi^{7/2}}{42} + \frac{\pi^{11/2}}{1320} - \frac{\pi^{15/2}}{75600} \].

Key concepts

Definite Integral

In the world of calculus, a definite integral is a fundamental concept used to calculate the area under a curve within a specific interval. In simple terms, if you imagine a curve on a graph, the definite integral helps you find the total space between that curve and the x-axis over a certain range. For instance, if we are given the integral \( \int_{a}^{b} f(x) \, dx \), it means we're calculating the area under the function \( f(x) \) from point \( a \) to point \( b \). The definite integral can be thought of as the accumulation of the quantities represented by the function in that interval. It's not just numbers, but the historical pile-up of the functions' values as they stretch along the x-axis. When you evaluate a definite integral, you often end up with a numerical value. This process is crucial in various fields, as it helps determine quantities like distances, areas, and even probabilities. In our exercise, we use the definite integral of the function \( \sin(x^2) \) from 0 to \( \sqrt{\pi} \) to find the area under this curve over the specified range.

Approximation Method

Sometimes, finding the exact value of a definite integral is challenging, especially when dealing with complex functions. This is where approximation methods come in handy. When the integral cannot be easily solved or expressed in a closed form, we turn to methods such as the Taylor series to estimate the value.The Taylor series allows us to express complicated functions as an infinite sum of terms calculated from the function's derivatives at a single point. This is particularly useful for trigonometric functions like sine, where the exact integral might be tricky to determine.In our exercise, we've used the Taylor series expansion to approximate \( \sin(x^2) \). By taking only the first four nonzero terms of this expansion, we simplify our calculations while still getting an approximation close enough for practical purposes. This approach illustrates the power and utility of approximation methods in mathematical analysis, especially when dealing with integrals of functions that aren't elementary.

Sin Function

The sine function, denoted as \( \sin(x) \), is one of the most fundamental trigonometric functions. It's widely known for its wave-like pattern, which repeats every \(2\pi\). This periodic nature makes it crucial in analyzing oscillatory motions, waves, and even in signal processing.In the context of our exercise, we specifically deal with \( \sin(x^2) \), a variant that modifies the typical sine wave pattern. By replacing \( x \) with \( x^2 \), the behavior of the function changes, elongating and stretching the sine wave. This variant is a bit challenging to integrate directly, hence the need for Taylor series expansion.The sine function's derivatives, alternating between sine and cosine functions, lend themselves well to substitution and expansion in Taylor series. This interplay of derivatives is what allows us to approximate \( \sin(x^2) \) effectively, simplifying the integration process by breaking it down into manageable polynomial components.

Integration of Series

Integration of series is a powerful technique that simplifies the process of integrating complex functions. When a function is expressed as a sum of its Taylor series, each term in the series can be integrated individually over the specified interval, and then summed up to find the total integral.This method is extremely useful because integrating polynomials—like the terms in the Taylor series—is generally straightforward and manageable. In the case of \( \sin(x^2) \), by expanding it into a series \( x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \frac{x^{14}}{5040} \), we transform the integration task into something much simpler.For our specific exercise, we integrate each of these polynomial terms from 0 to \( \sqrt{\pi} \), efficiently evaluating the resulting integral for the function. The sum of these evaluated integrals approximates the definite integral of the original function, exemplifying how breaking down a complex task into simpler parts can yield practical results, especially in approximation problems.