Problem 51 - Applied Calculus

Question

Use integration by substitution and the Fundamental Theorem to evaluate the definite integrals. $$\int_{0}^{3} \frac{2 x}{x^{2}+1} d x$$

Step-by-Step Solution

Answer

The integral evaluates to \(\ln(10)\).

Step 1: Identify the Substitution Variable

We need to simplify the integral by choosing an appropriate substitution. Observe that the derivative of \(x^2 + 1\) is \(2x\), which is present in the numerator. Thus, let's set \(u = x^2 + 1\).

Step 2: Differentiate the Substitution Variable

Differentiate the substitution variable both with respect to \(x\), giving \(\frac{du}{dx} = 2x\). Therefore, \(du = 2x \, dx\).

Step 3: Substitute in the Integral

Using \(u = x^2 + 1\) and \(du = 2x \, dx\), the integral becomes \[\int \frac{2x}{x^2+1} \, dx = \int \frac{1}{u} \, du.\]

Step 4: Evaluate the Indefinite Integral

The integral \(\int \frac{1}{u} \, du = \ln |u| + C\). So, the antiderivative becomes \(\ln |x^2+1| + C\).

Step 5: Apply the Limits of Integration

Replace \(u\) with \(x^2 + 1\) in the indefinite integral. Now, compute the definite integral from 0 to 3:\[\left[ \ln (x^2 + 1) \right]_0^3 = \ln (3^2 + 1) - \ln (0^2 + 1).\]

Step 6: Calculate the Final Values

Substitute the values into the antiderivative expression, giving:\[\ln (9 + 1) - \ln (0 + 1) = \ln (10) - \ln (1).\]Since \(\ln (1) = 0\), the result becomes \(\ln (10)\).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with that of integration. It is a crucial theorem in calculus providing a way to compute definite integrals. This theorem consists of two main parts. The first part states that if a function is continuous on a closed interval \([a, b]\), then it has an antiderivative over that interval. The second part tells us how to evaluate a definite integral by using this antiderivative.
The second part (often the more used part in integral calculus) states that if \(F\) is an antiderivative of a function \(f\) over an interval \( [a, b]\), then:

• \( \int_a^b f(x) \, dx = F(b) - F(a) \)

This means we first find one antiderivative of the function we want to integrate, then evaluate it at the upper and lower limits of integration, and finally subtract the two results. In the example given, after performing the substitution, we used the theorem to compute the definite integral from 0 to 3 using the antiderivative \( \ln(x^2 + 1) \). By evaluating it at the endpoints, we found the integral.

definite integrals

A definite integral is a type of integral that calculates the accumulation of quantities, such as total distance or total area, over a specific interval. Unlike indefinite integrals, which yield a function, definite integrals provide a numeric result. The concept involves integrating a function within specified limits, usually denoted as the lower and upper limits of integration. For the expression \( \int_a^b f(x) \, dx \), \(a\) represents the lower limit, and \(b\) is the upper limit.
The process involves several steps:

• Determine an antiderivative \(F(x)\) for the function \(f(x)\).
• Evaluate \(F(b)\) and \(F(a)\).
• Subtract these to produce the result, \(F(b) - F(a)\).

When solving the original exercise, the definite integral \( \int_0^3 \frac{2x}{x^2+1} \, dx \) was calculated by substituting and then integrating, producing the value \( \ln(10) \). Definite integrals are often applied in physics and statistics to calculate total quantities.

substitution method

The substitution method is a technique used in integral calculus to simplify complex integrals by changing variables. Often, an integral may contain a composition of functions that is hard to integrate directly. In such cases, substitution can transform the integral into a simpler form.
Here's how the substitution method generally works:

• Identify part of the integrand that complicates the integration, often spotting a derivative in the expression.
• Set a substitution variable \(u\), often related to this part, making the integral easier to work with.
• Rewrite the integral in terms of \(u\) and \(du\).

As shown in the original exercise, the substitution \(u = x^2 + 1\) was chosen because its derivative, \(2x\), was also present in the integrand. This allowed for the transformation \(du = 2x \, dx\), simplifying the given integral into \(\int \frac{1}{u} \, du\). Such transformations enable us to leverage simpler integral forms, making them solvable with ease.