Problem 83 - Thomas Calculus

Question

Suppose that \(F(x)\) is an antiderivative of \(f(x)=(\sin x) / x, x>0\) Express $$\int_{1}^{3} \frac{\sin 2 x}{x} d x$$ in terms of \(F .\)

Step-by-Step Solution

Answer

\(\frac{1}{2} ( F(6) - F(2) )\).

Step 1: Substitution Identification

To express \(\int_{1}^{3} \frac{\sin 2x}{x} \, dx\) in terms of \(F(x)\), notice the integral involves \(\sin 2x\). We need to substitute \(u = 2x\), making the substitution easier as it involves transforming the limits and the function.

Step 2: Calculate Differential

Calculate the differential of \(u\). If \(u = 2x\), then \(du = 2\,dx\). Solving for \(dx\), we find \(dx = \frac{1}{2} \, du\).

Step 3: Substitution in Integral

Substitute \(u = 2x\) and \(dx = \frac{1}{2} \, du\) into the original integral. The original integral becomes:\[\int_{1}^{3} \frac{\sin 2x}{x} \, dx = \int_{2}^{6} \frac{\sin u}{u/2} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_{2}^{6} \frac{\sin u}{u} \, du.\]

Step 4: Adjust Limits of Integration

When \(x = 1\), \(u = 2(1) = 2\);and when \(x = 3\), \(u = 2(3) = 6\). So the limits of integration change accordingly.

Step 5: Express in Terms of Given Antiderivative

Recognizing that \(F(u)\) is the antiderivative of \(\frac{\sin u}{u}\), we can write the integral as:\[\frac{1}{2} \left[ F(u) \right]_{2}^{6} = \frac{1}{2} ( F(6) - F(2) ).\] This step uses the Fundamental Theorem of Calculus to evaluate the definite integral.

Key Concepts

Substitution Method

The substitution method is a powerful technique used in calculus to solve integrals. By substituting a part of the integral with a new variable, we simplify the integration process. This is especially useful when dealing with complex functions. In this exercise, we use substitution by setting \( u = 2x \). This change simplifies the function inside the integral, turning \( \sin 2x \) into \( \sin u \).
This substitution simplifies the integral, making it easier to evaluate. By changing \( dx \) to \( du \) using the relation \( du = 2 \, dx \), which means \( dx = \frac{1}{2} \, du \), we are able to substitute everything in terms of \( u \) and \( du \). This approach changes both the function you're integrating and the limits of integration, resulting in a more manageable integrand.

Definite Integral

A definite integral represents the area under a curve between two points on the x-axis. In our example, we calculate \( \int_{1}^{3} \frac{\sin 2x}{x} \, dx \), which is the area between the curve \( \frac{\sin 2x}{x} \) from \( x = 1 \) to \( x = 3 \).

Definite integrals are closed intervals, resulting in a single numerical value. Unlike indefinite integrals, they include limits of integration, which means the interval is bounded. After substitution, the limits of our integral change from \( x \) values 1 to 3, to \( u \) values 2 to 6, reflecting our substitution \( u = 2x \). This adjustment ensures that the definite integral accurately measures the desired area under the curve.

Antiderivative

An antiderivative, sometimes known as an indefinite integral, of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \). In simple terms, it is a function whose derivative returns the original function. In our case, we know \( F(x) \) is an antiderivative of \( \frac{\sin x}{x} \), meaning that if you differentiate \( F(x) \), you would obtain \( \frac{\sin x}{x} \).
To express a solution by involvement of an antiderivative, we use the properties of the antiderivative to evaluate definite integrals. In our situation, once we have changed the integral with substitution, having the integral of \( \frac{\sin u}{u} \) allows the use of the antiderivative \( F(u) \) to express the answer in terms of \( F(6) \) and \( F(2) \).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, two key operations in calculus. This theorem has two parts, with the relevant part to our problem stating that if \( F(x) \) is an antiderivative of \( f(x) \), then the definite integral of \( f(x) \) over \([a, b] \) is \( F(b) - F(a) \).

In our example, this allows us to evaluate \( \frac{1}{2} \int_{2}^{6} \frac{\sin u}{u} \, du \) as \( \frac{1}{2}[F(6) - F(2)] \). By subtracting the values of \( F(u) \) at these limits, we get the definite integral's value. This shows how the Fundamental Theorem of Calculus makes evaluating definite integrals feasible through antiderivatives.