Problem 21 - Applied Calculus

Question

Use substitution to express each of the following integrals as a multiple of \(\int_{a}^{b}(1 / w) d w\) for some \(a\) and \(b\). Then evaluate the integrals. (a) \(\int_{0}^{1} \frac{x}{1+x^{2}} d x\) (b) \(\int_{0}^{\pi / 4} \frac{\sin x}{\cos x} d x\)

Step-by-Step Solution

Answer

(a) \( \frac{1}{2} \ln 2 \); (b) \( \frac{1}{2} \ln 2 \).

Step 1: Recognize the Appropriate Substitution for Part (a)

We are given the integral \( \int_{0}^{1} \frac{x}{1+x^{2}} dx \). Notice that the derivative of \( 1 + x^2 \) is \( 2x \), which suggests a substitution \( u = 1 + x^2 \). This implies \( du = 2x \, dx \) or \( x \, dx = \frac{1}{2} du \). If we substitute: when \( x = 0 \), \( u = 1 \) and when \( x = 1 \), \( u = 2 \). Thus, the integral becomes \( \int_{1}^{2} \frac{1}{2u} du \).

Step 2: Integral Transformation for Part (a)

We now have \( \int_{1}^{2} \frac{1}{2u} du \), which can be factored as \( \frac{1}{2} \int_{1}^{2} \frac{1}{u} du \). Observe that this is a multiple of \( \int_{a}^{b} \frac{1}{w} dw \) with \( a = 1 \) and \( b = 2 \).

Step 3: Evaluate the Integral for Part (a)

Evaluate \( \int_{1}^{2} \frac{1}{u} du = [\ln |u|]_{1}^{2} = \ln 2 - \ln 1 = \ln 2 \). Multiply by \( \frac{1}{2} \): \( \frac{1}{2} \ln 2 \). So, the evaluated integral is \( \frac{1}{2} \ln 2 \).

Step 4: Recognize the Appropriate Substitution for Part (b)

The integral given is \( \int_{0}^{\pi/4} \frac{\sin x}{\cos x} dx \). Recognize this as \( \int_{0}^{\pi/4} \tan x \, dx \). Use the substitution \( u = \cos x \), which gives \( du = -\sin x \, dx \) or \( \sin x \, dx = -du \). When \( x = 0 \), \( u = 1 \) and when \( x = \pi/4 \), \( u = \frac{\sqrt{2}}{2} \). Thus, the integral becomes \( -\int_{1}^{\frac{\sqrt{2}}{2}} \frac{1}{u} du \).

Step 5: Integral Transformation for Part (b)

The transformed integral is \(-\int_{1}^{\frac{\sqrt{2}}{2}} \frac{1}{u} du \), which is equivalent to a negative multiple of \( \int_{a}^{b} \frac{1}{w} dw \) where \( a = 1 \) and \( b = \frac{\sqrt{2}}{2} \).

Step 6: Evaluate the Integral for Part (b)

Evaluate \( -\int_{1}^{\frac{\sqrt{2}}{2}} \frac{1}{u} du = -([\ln |u|]_{1}^{\frac{\sqrt{2}}{2}}) = - \left( \ln \left( \frac{\sqrt{2}}{2} \right) - \ln 1 \right) \). Simplifying gives: \( \ln 2^{-1/2} = -\frac{1}{2} \ln 2 \). Therefore, the evaluated integral is \( \frac{1}{2} \ln 2 \).

Key Concepts

Integrals

Integrals are a fundamental concept in calculus that allow us to determine the area under a curve. They serve as the opposite process of differentiation. One core purpose of integration is to solve problems involving continuously accumulating quantities.

• Indefinite Integrals: These are expressed without specific numerical limits, denoting an entire family of functions as antiderivatives.
• Definite Integrals: These are calculated over a specific interval and yield a numerical value, representing the area under the curve of a function within that interval.

The process of finding integrals can sometimes be challenging depending on the function involved. In such cases, substitution methods can simplify the task. This technique involves changing the variable of integration to make the integral easier to evaluate. It's analogous to the chain rule for differentiation, where you're essentially undoing a complex composite function.

Definite Integrals

Definite integrals, as mentioned earlier, compute the area under a specific segment of a curve. They are notated as \[ \int_{a}^{b} f(x) \, dx \] where \(a\) and \(b\) are the limits of integration. Definite integrals have some distinct characteristics:

• Defined Limits: The integration takes place between two points, yielding a fixed value.
• Net Area: It measures the signed area, meaning it considers both positive and negative areas under the curve depending on crossing axes.

In exercises involving definite integrals, boundary values are transformed to match the new variable when substitution is used. This ensures that the limits reflect the updated context of the integral, as seen in the problem exercises provided.
Using the substitution method, the problem aimed to convert a relatively complex function into a simpler form by redefining the variable and adjusting integration limits accordingly. This made the evaluation process straightforward and computationally feasible.

Trigonometric Substitution

For certain integrals, especially those involving trigonometric functions, trigonometric substitution can be used to simplify the integral. The idea is to replace the given variable with a trigonometric function that simplifies the expression, often used when dealing with forms like \( a^2 - x^2 \), \( a^2 + x^2 \), or \( x^2 - a^2 \).
The exercise involving \[ \int_{0}^{\pi/4} \frac{\sin x}{\cos x} dx \] is simplified through this method. Here, recognizing that \( \frac{\sin x}{\cos x} \) is \( \tan x \) allowed us to use trigonometric identities to transform the integration process.

• Identify Substitution: Choose a function that can represent the original variables in terms of a trigonometric identity.
• Adjust Limits: When integrating, adjust the limits to correspond with the trigonometric substitution made.
• Simplify and Solve: The integral often becomes easier to evaluate, sometimes turning into a form that is recognizable and straightforward to integrate.

Trigonometric substitution not only aids in solving integrals but also offers a deeper understanding of the relationships within trigonometric identities.