Problem 49 - Calculus Early Transcendentals

Question

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral \(\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}\). Answer: The definite integral \(\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}\) is equal to \(\frac{\pi}{12}\).

Step 1: Find the antiderivative of the given function

From the given exercise, the function to integrate is \(\frac{1}{1+x^2}\). The antiderivative of this function is the arctangent function, so the indefinite integral is: $$F(x) = \arctan(x) + C$$ where \(C\) is an arbitrary constant.

Step 2: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, the result of the definite integral is the difference between the antiderivative evaluated at the upper and lower limits of integration. Therefore, we need to find \(F(\sqrt{3})\) and \(F(1)\) and subtract their values to obtain the result: $$F(\sqrt{3}) = \arctan(\sqrt{3})$$ and: $$F(1) = \arctan(1)$$

Step 3: Subtract the values to find the definite integral

To find the value of the definite integral, subtract the values obtained in Step 2: $$\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}} = F(\sqrt{3}) - F(1) = \arctan(\sqrt{3}) - \arctan(1) = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$$ So the definite integral \(\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}\) is equal to \(\frac{\pi}{12}\).

Key Concepts

Definite Integrals

When we talk about definite integrals, we're looking at a core concept in calculus that quantifies the accumulation of quantities, such as area under a curve, between two points on a graph. The definite integral is represented as an integral with specific boundaries, generally noted as \( a \) and \( b \) for the lower and upper limits, respectively.

In mathematical terms, the definite integral of a function \( f(x) \) from \( a \) to \( b \) is written as \( \int_a^b f(x) \,dx \). This notation denotes the sum of the infinitesimal slices or areas under the curve of \( f(x) \) from \( a \) to \( b \) along the x-axis.

To calculate a definite integral, you would typically find the antiderivative of the function—which acts as a backward operation to differentiation—and then apply the limits of integration. Doing this gives you the net area enclosed by the function within the given range.

Antiderivative

An antiderivative, also known as an indefinite integral, is essentially the reverse of a derivative. It refers to the process of finding a function \( F(x) \) whose derivative is the given function \( f(x) \). In other words, if \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).

There can be infinitely many antiderivatives for a given function, since adding any constant \( C \) to an antiderivative still results in a valid antiderivative. This constant of integration, \( C \) is what distinguishes indefinite integrals from definite integrals, which have fixed limits and do not include the constant. The notation \( \int f(x) \,dx = F(x) + C \) is used to express the most general antiderivative of \( f(x) \).

Arctangent Function

The arctangent function, denoted as \( \arctan(x) \), is the inverse of the tangent function. It is used to find an angle whose tangent value is \( x \). This function is particularly important when working with trigonometric integrals, especially when the integrand involves \( \frac{1}{1+x^2} \).

Since the derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \), it is common to encounter \( \arctan(x) \) as the antiderivative when integrating functions of this form. Understanding the properties of the arctangent function, such as its range being between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), is critical for solving related integrals accurately.

Integration Limits

Integration limits are the boundaries that specify the interval over which an integral is to be calculated. These limits are essential when determining the value of a definite integral. They can be either numbers, representing points on the x-axis, or expressions which might also involve variables or even infinity, indicating an unbounded region.

In the given exercise, we used 1 and \( \sqrt{3} \) as the lower and upper limits for the integral of \( \frac{1}{1+x^2} \), respectively. The use of limits allows us to compute the exact value of an area or an accumulated quantity within a specific range on the graph of a function. Proper selection and understanding of integration limits are key for correctly interpreting the results of a definite integral.