Problem 109 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{0}^{1} \frac{1}{1+x^{2}} d x $$

Step-by-Step Solution

Answer

The value of the definite integral is \( \frac{\pi}{4} \).

Step 1: Recognize the Integral Form

The integral given is \[\int_{0}^{1} \frac{1}{1+x^{2}} dx\] which is a standard form recognizing as the derivative of the inverse tangent function, \( \tan^{-1}(x) \).

Step 2: Find the Antiderivative

The antiderivative of \( \frac{1}{1+x^{2}} \) is \( \tan^{-1}(x) \). Therefore, the indefinite integral is\[\int \frac{1}{1+x^{2}} dx = \tan^{-1}(x) + C\] where \( C \) is the constant of integration.

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, evaluate the antiderivative at the upper limit of 1 and subtract its evaluation at the lower limit of 0:\[\int_{0}^{1} \frac{1}{1+x^{2}} dx = \tan^{-1}(1) - \tan^{-1}(0)\]

Step 4: Evaluate the Expression

Calculate each inverse tangent value.\( \tan^{-1}(1) \) is \( \frac{\pi}{4} \) because the tangent of \( \frac{\pi}{4} \) is 1.\( \tan^{-1}(0) \) is 0 because the tangent of 0 is 0.Substitute these into the expression:\[\frac{\pi}{4} - 0 = \frac{\pi}{4}\]

Key Concepts

Inverse Trigonometric Functions

In mathematics, inverse trigonometric functions are the inverse operations of the basic trigonometric functions like sine, cosine, and tangent. These inverses are crucial since they help us solve equations that involve trigonometric functions.
For the tangent function, its inverse is denoted as \tan^{-1}(x)\ or sometimes as \text{arctan}(x)\. This inverse function helps us determine the angle whose tangent is a given number.

• For example, if \tan^{-1}(1)\, it means we are looking for the angle whose tangent is 1, which is \(\frac{\pi}{4}\).
• Similarly, \tan^{-1}(0)\ means finding the angle whose tangent is 0, resulting in 0.

These functions are integrable in calculus, and recognizing forms like \frac{1}{1+x^2}\ as \frac{d}{dx}[\tan^{-1}(x)]\ is vital in solving integrals that involve these forms.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone in the field of calculus. It creates a connection between differentiation and integration, the two primary operations in calculus. This theorem is made up of two main parts.
One part states that if you have a continuous function, then its integral over an interval can be calculated using its antiderivative. In simpler terms, to evaluate a definite integral, you find the antiderivative of the function and compute its difference at the upper and lower bounds.

• For instance, to evaluate \(\int_{0}^{1} \frac{1}{1+x^2}\, dx\), you first determine the antiderivative, which here is \tan^{-1}(x)\.
• Next, you calculate this at the boundaries: \tan^{-1}(1)\ and \tan^{-1}(0)\, leading to the answer \(\frac{\pi}{4}\).

Thus, the theorem seamlessly bridges the process of finding areas under a curve (integration) with the derivative operations.

Antiderivatives

An antiderivative, also known as an indefinite integral, is essentially the reverse of taking a derivative. Given a function, its antiderivative is another function whose derivative is the original function.
Finding the antiderivative is a crucial step when solving integrals. For determining definite integrals, knowing the antiderivative allows us to apply the Fundamental Theorem of Calculus effectively.

• For example, in the integral \(\int \frac{1}{1+x^{2}}\, dx\), the antiderivative is \tan^{-1}(x)\ because taking the derivative of \tan^{-1}(x)\ yields \frac{1}{1+x^2}\.
• This means the indefinite integral can be written as \(\tan^{-1}(x) + C\), where \C\ is a constant used to account for all possible antiderivatives.

By recognizing standard forms and their antiderivatives, computation of integrals becomes much more straightforward.