Question

Evaluate the definite integral \(\int_{\sqrt{3} / 3}^{\sqrt{3}} \frac{d x}{1+x^{2}}\).

Short answer

The integral evaluates to \( \frac{\pi}{6} \).

Step-by-step solution

Identify the Integral

We are tasked with evaluating the definite integral \( \int_{\sqrt{3}/3}^{\sqrt{3}} \frac{dx}{1+x^{2}} \). Recognize that the integrand \( \frac{1}{1+x^2} \) is the derivative of the arctangent function.

Apply the Fundamental Theorem of Calculus

Since \( \frac{1}{1+x^2} \) is the derivative of \( \arctan(x) \), the integral evaluates to \( \arctan(x) \). According to the Fundamental Theorem of Calculus, we evaluate this antiderivative from \( \sqrt{3}/3 \) to \( \sqrt{3} \).

Evaluate the Antiderivative at the Upper Limit

First evaluate \( \arctan(x) \) at the upper limit \( x = \sqrt{3} \). We find \( \arctan(\sqrt{3}) = \frac{\pi}{3} \), since \( \tan(\frac{\pi}{3}) = \sqrt{3} \).

Evaluate the Antiderivative at the Lower Limit

Next, evaluate \( \arctan(x) \) at the lower limit \( x = \sqrt{3}/3 \). We find \( \arctan(\sqrt{3}/3) = \frac{\pi}{6} \), since \( \tan(\frac{\pi}{6}) = \sqrt{3}/3 \).

Subtract the Results

Subtract the antiderivative evaluated at the lower limit from the antiderivative evaluated at the upper limit: \[ \int_{\sqrt{3}/3}^{\sqrt{3}} \frac{dx}{1+x^{2}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}. \] This is the value of the definite integral.

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is a powerful bridge connecting differentiation and integration, two essential operations in calculus. The theorem tells us two main things:

• First, it provides a way to evaluate definite integrals using antiderivatives.
• Second, it ensures that every continuous function has an antiderivative.

This theorem states that if a function is continuous on a closed interval, then the function's definite integral can be evaluated using its antiderivative at the endpoints of the interval.
In essence, if you have the definite integral \[\int_{a}^{b} f(x)\,dx\]and you know an antiderivative \(F(x)\) of \(f(x)\), then: \[\int_{a}^{b} f(x)\,dx = F(b) - F(a)\]
For instance, in our exercise, we evaluate \( \int_{\sqrt{3}/3}^{\sqrt{3}} \frac{dx}{1+x^{2}} \)by recognizing that an antiderivative of \(\frac{1}{1+x^{2}}\) is \(\arctan(x)\) and evaluating it at the endpoints \( \sqrt{3}/3 \) and \( \sqrt{3} \).
This leads to finding the values \(\arctan(\sqrt{3})\) and \(\arctan(\sqrt{3}/3)\) and subtracting them, yielding the result of the integral.

arctangent function

The arctangent function, often denoted as \(\arctan(x)\) or \(\tan^{-1}(x)\), is the inverse function of the tangent trigonometric function. It is crucial in various applications, especially when dealing with triangles and angles.
The arctangent function provides the angle whose tangent is a given number.
For example:

• \(\arctan(\sqrt{3}) = \frac{\pi}{3}\) because \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\).
• Similarly, \(\arctan\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}\) since \(\tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}\).

Understanding these fundamental trigonometric identities helps in solving integrals where arctangent appears as the antiderivative, like in integral exercises similar to ours.
The range of the arctangent function is \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), ensuring that every real number has a corresponding angle within this interval.

antiderivative evaluation

Antiderivative evaluation is a critical step when applying the Fundamental Theorem of Calculus to solve definite integrals. It involves finding an antiderivative for the integrand, evaluating the antiderivative at the given limits, and then computing the difference.
Let's break this process down:

• Finding the Antiderivative: Identify a function whose derivative matches the integrand. For example, the antiderivative of \(\frac{1}{1+x^{2}}\) is \(\arctan(x)\).
• Evaluate at Limits: Calculate the antiderivative at the upper and lower limits of the integral. Given the exercise, this meant finding \(\arctan(\sqrt{3})\) and \(\arctan(\sqrt{3}/3)\).
• Compute the Difference: Finally, subtract the value of the antiderivative at the lower limit from its value at the upper limit. This subtraction gives the value of the definite integral, which was \(\frac{\pi}{6}\) in our situation.

Breaking down the problem into these three manageable steps can simplify the often challenging task of evaluating definite integrals.