Problem 30 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^{1}_{0} \frac{4}{1 + p^2} \,dp \)

Step-by-Step Solution

Answer

The value of the integral is \( \pi \).

Step 1: Recognize the integral formula

The integral \( \int \frac{1}{1 + p^2} \,dp \) corresponds to the well-known inverse tangent function. This is because the derivative of \( \tan^{-1}(p) \) is \( \frac{1}{1 + p^2} \).

Step 2: Apply the antiderivative

Using the antiderivative we recognized in Step 1, the integral becomes \( 4 \cdot \int \frac{1}{1 + p^2} \,dp = 4 \cdot \tan^{-1}(p) + C \). Here, \( C \) is the constant of integration.

Step 3: Evaluate the definite integral

We need to evaluate the definite integral from 0 to 1. Thus, we compute \( 4 \cdot \left[ \tan^{-1}(p) \right]_{0}^{1} \).

Step 4: Substitute the limits of integration

Substitute the upper limit: \( \tan^{-1}(1) = \frac{\pi}{4} \). Substitute the lower limit: \( \tan^{-1}(0) = 0 \).

Step 5: Compute the result

Calculate the result: \( 4 \cdot \left( \frac{\pi}{4} - 0 \right) = \pi \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential tools in calculus, especially when working with integrals that involve expressions resembling trigonometric derivatives. The most common inverse trigonometric functions include arcsine \(\( \sin^{-1}(x) \)\), arccosine \(\( \cos^{-1}(x) \)\), and arctangent \(\( \tan^{-1}(x) \)\), among others.

These functions help us find angles for given trigonometric values. More importantly in calculus, they allow us to identify and integrate certain forms. For instance, recognizing that the derivative of \(\( \tan^{-1}(p) \)\) is \(\( \frac{1}{1 + p^2} \)\), as used in the original exercise, enables the evaluation of integral expressions like \(\( \int \frac{1}{1 + p^2} \mathrm{d}p \)\) using the arctangent function.

• The characteristic shape of \(\( \int \frac{1}{1 + p^2} \mathrm{d}p \)\) suggests the arctangent's involvement.
• These functions help simplify complex integrals by relating them to trigonometric identities.

Understanding these connections aids in solving integrals efficiently, paving the way for deeper explorations in calculus.

Integration Techniques

Integration techniques provide crucial methods for solving various types of integrals. Unlike differentiation, which is rather straightforward, integration often requires finesse and creativity. Recognizing when to use basic integration rules or special techniques is vital.

For instance, you might encounter integrals that fit the form of known derivatives—these can be tackled using direct antiderivatives. In the original exercise, we leveraged the fact that the integral of \(\( \frac{1}{1 + p^2} \)\) yields \(\( \tan^{-1}(p) \)\), multiplied by the constant 4.

• Consider substitution or partial fraction decomposition when straightforward antiderivatives aren't evident.
• Use trigonometric identities to simplify or restructure integrals, especially those involving \(\( \sqrt{1-x^2} \)\) or similar terms.

Mastering these techniques opens up a variety of pathways to solve seemingly complex problems, turning challenges into manageable steps.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of mathematical analysis, creating a bridge between differentiation and integration. It reveals that differentiation and integration are inverse processes.

This theorem consists of two main parts. The first part states that if a function is continuous over a closed interval \(\( [a, b] \)\), then the integral of the function over that interval is connected to its antiderivative. The theorem ensures that we can calculate definite integrals using antiderivatives found through indefinite integrals.

• The second part asserts that the derivative of an integral function returns the original function, under the right conditions.
• Applied in definite integrals, such as in our example, it allows us to calculate the area under curves precisely.

By leveraging this theorem, we can compute integrals like \(\( \int_{0}^{1} \frac{4}{1+p^2} \mathrm{d}p \)\) with ease, using the limits of integration and the antiderivative calculated.