Problem 67 - Calculus for Biology and Medicine

Question

Evaluate each definite integral. $$ \int_{-1}^{0} \frac{2}{1+x^{2}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{2} \).

Step 1: Identify the Integral Type

The integral to evaluate is \( \int_{-1}^{0} \frac{2}{1+x^{2}} \, dx \). It is a definite integral of a function in the form of \( \frac{2}{1+x^2} \), which resembles the derivative of an inverse trigonometric function.

Step 2: Recall the Antiderivative

Recall that \( \frac{1}{1+x^2} \) is the derivative of the arctangent function \( \tan^{-1}(x) \). Therefore, the antiderivative of \( \frac{2}{1+x^2} \) is \( 2 \tan^{-1}(x) \).

Step 3: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F \) is an antiderivative of \( f \) on an interval, then \( \int_a^b f(x) \, dx = F(b) - F(a) \). For our problem, set \( F(x) = 2 \tan^{-1}(x) \).

Step 4: Evaluate the Antiderivative at the Bounds

Compute \( F(0) \) and \( F(-1) \):\[ F(0) = 2 \tan^{-1}(0) = 2 \times 0 = 0 \] \[ F(-1) = 2 \tan^{-1}(-1) = 2 \times \left( -\frac{\pi}{4} \right) = -\frac{\pi}{2} \].

Step 5: Calculate the Definite Integral

The value of the definite integral is \( F(0) - F(-1) \): \[ 0 - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2} \].

Key Concepts

Antiderivative

In calculus, finding the antiderivative, also known as the indefinite integral, refers to determining a function whose derivative is the given function. This is the reverse process of differentiation. For example, if you know that the derivative of the function \( f(x) \) is \( g(x) \), then \( f(x) \) is an antiderivative of \( g(x) \).

In our exercise, the function \( \frac{2}{1+x^2} \) was identified to closely match the derivative of the inverse trigonometric function \( \tan^{-1}(x) \). Hence, its antiderivative is \( 2\tan^{-1}(x) \). Recognizing the form of functions relative to known derivatives is a key skill in solving integration problems.

• The prefix "anti" suggests it is the opposite of finding the derivative.
• Once we know the basic antiderivative, we can apply it directly in definite integral evaluations.
• This process is crucial in integrating functions that may not already appear in standard forms.

Having a solid understanding of basic antiderivatives will make evaluating more complex integrals less challenging.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concept of differentiation with integration. It has two primary parts. The first part tells us that integration can be reversed by differentiation, while the second part provides a practical way to compute definite integrals.

For the integral \( \int_a^b f(x) \, dx \), the theorem states that if \( F \) is an antiderivative of \( f \) over an interval, then the integral from \( a \) to \( b \) can be calculated as \( F(b) - F(a) \).

• It simplifies the process of finding the signed area under a curve.
• Applies to continuous functions over a closed interval.
• It allows us to evaluate complex integrals by finding an antiderivative and using simple arithmetic.

In our exercise, this theorem was used to compute \( \int_{-1}^{0} \frac{2}{1+x^2} \, dx \) by evaluating \( F(x) = 2 \tan^{-1}(x) \) at the limits \( 0 \) and \( -1 \). This underscores the theorem’s utility in connecting antiderivatives to calculating definite areas.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in both differentiation and integration. They allow us to find angles given trigonometric ratios, and are crucial in calculus for their derivatives, which often appear in integration problems.

Some common inverse trigonometric functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \). Each has its own derivative:

• \( \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2} \)
• \( \frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}} \)
• \( \frac{d}{dx} \cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}} \)

Understanding these functions' derivatives is key when you encounter integrals resembling these forms. They show up frequently in real-world problems where angles and movement around circles or rotations are involved.

In particular, the exercise involved recognizing that \( \frac{1}{1+x^2} \) is the derivative of \( \tan^{-1}(x) \), which helped determine the antiderivative. Thus, familiarity with inverse trigonometric functions expands our toolset significantly when solving integrals.