Problem 29 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Evaluate the integral. In many cases it will be advantageous to begin by doing a substitution. For example, in Problem 19, let \(w=\sqrt{x}, w^{2}=x ;\) then \(2 w d w=d x .\) This eliminates \(\sqrt{x}\) by replacing \(x\) with a perfect square. $$ \int_{0}^{2} e^{\sqrt{x}} d x $$

Step-by-Step Solution

Answer

The value of the integral is: \(2e^{\sqrt{2}} - 4\)

Step 1: Set up the substitution

First, perform a substitution. Let \(u = \sqrt{x}\). Therefore, \(u^2 = x\). Differentiating both sides gives \(2u du = dx\). Also make necessary change in the bounds. When \(x=0\), \(u=0\) and when \(x=2\), \(u=\sqrt{2}\).

Step 2: Substitute into the integral

Substitute all the \(x\) and \(dx\) terms in the integral with the terms we have gotten from our substitution, giving: \[ \int_{0}^{\sqrt{2}} e^{u} 2u du \]

Step 3: Solve the Integral

This integral is a perfect form for integration by a method called integration by parts, with \(v = e^{u}\) and \(du = 2u\). Apply the integration by parts formula \[ \int udv = uv- \int vdu\]. This results in \(2[u e^u - \int e^u du] \]

Step 4: Evaluate the integral and resubstitute

The result after integrating and simplifying is \(2[ue^u - e^u]\) evaluated between \(0\) and \(\sqrt{2}\). Resubstitute back \(u = \sqrt{x}\) and compute the definite integral.

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus that represents the accumulation of quantities and allows us to calculate things like areas under curves. To understand a definite integral, visualize the area between the function and the x-axis as being filled with an infinite number of infinitely thin rectangles. The process of integration sums up the areas of these rectangles to find a total area.

For example, the definite integral from a to b, represented as \( \int_{a}^{b} f(x) dx \), computes the total area under the curve f(x) from x=a to x=b. In the given exercise, a definite integral of the function \( e^{\sqrt{x}} \) from 0 to 2 is required. This is an example of how calculus is used to find the accumulated sum of continuous quantities over a specified interval.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It's based on the product rule for differentiation and can be used to integrate more complex functions that cannot be integrated using standard methods.

The formula for integration by parts is \( \int u dv = uv - \int v du \), where u and v are functions of x. The challenge lies in choosing the appropriate functions for u and dv to simplify the integral. In our exercise, the technique is suggested in Step 3 after substituting, but it should be noted that a more fitting approach after substitution would be a straightforward integration without the need for integration by parts, as the resulting integral is simpler than initially anticipated.

U-Substitution

U-Substitution, also known as variable substitution, is a method for finding integrals that can transform a difficult integral into a simpler one. It works by choosing a substitution that simplifies the integrand.

In our step-by-step solution, u-substitution is applied by setting \( u = \sqrt{x} \) to transform the integral of \( e^{\sqrt{x}} \) into something more manageable. After determining \( u \) and \( du \) and substituting these into the integral, the limits of integration must also be changed to match the new variable, transforming the problem into its u-equivalent. However, the exercise solution takes an extra non-essential step by introducing integration by parts even after a successful substitution---a common mistake to watch out for.

Calculus

Calculus, the branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series, is an indispensable tool for understanding changes between values that are related by a function. Its two main types are differential calculus, which concerns the rate of change of quantities, and integral calculus, which focuses on the accumulation of quantities.

In the context of the given problem, integral calculus is used to understand the behavior of the function \( e^{\sqrt{x}} \) within a certain interval. Calculus methods, like u-substitution, allow us to make complex problems tractable and showcase the beauty and utility of this mathematical field.