Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

By interpreting each of the following integrals as an area and then calculating this area by a \(y\) -integration, evaluate: (a) \(\int_{0}^{1} \sqrt{\frac{1-x}{x}} d x\) (b) \(\int_{-1}^{1} \sqrt{\frac{1+x}{1-x}} d x\)

Step-by-Step Solution

Answer

(a) 1, (b) \( \pi \).

Step 1: Identify the region for part (a)

The integrand for part (a) is \( \sqrt{\frac{1-x}{x}} \), and the region of integration is \( x \) from 0 to 1. We interpret this as the area under the curve \( y = \sqrt{\frac{1-x}{x}} \) between \( x = 0 \) and \( x = 1 \).

Step 2: Change of variables for part (a)

To solve the integral, we use a substitution to express \( x \) in terms of \( y \). Let \( x = \frac{1}{1+y^2} \), which gives \( y = \sqrt{\frac{1-x}{x}} \). Then, differentiate to find \( dx = -\frac{2y}{(1+y^2)^2} dy \).

Step 3: Rewrite and evaluate integral for part (a)

With the substitution, the limits of integration change from \( x=0 \) to \( y=\infty \) and \( x=1 \) to \( y=0 \). The integral becomes \( \int_{\infty}^{0} -2y \cdot \frac{2y}{(1+y^2)^2} dy \), which simplifies to \( 2 \int_{0}^{\infty} \frac{y}{1+y^2} dy \). This integral can be evaluated as \( 2[\frac{1}{2}\ln(1+y^2)]_{0}^{\infty} \) resulting in 1.

Step 4: Identify the region for part (b)

The integrand for part (b) is \( \sqrt{\frac{1+x}{1-x}} \) with limits \( x \) from -1 to 1. We interpret this as the area under the curve \( y = \sqrt{\frac{1+x}{1-x}} \) between \( x = -1 \) and \( x = 1 \).

Step 5: Change of variables for part (b)

For part (b), use the substitution \( x = \cos(\theta) \), giving \( dx = -\sin(\theta) d\theta \). Substituting into the integral, we express \( y = \sqrt{\frac{1+x}{1-x}} = \tan(\frac{\theta}{2}) \).

Step 6: Evaluate the integral for part (b)

The new limits are \( \theta = 0 \) to \( \theta = \pi \). Substitute to get \( \int_{0}^{\pi} \sin(\theta) \cdot \tan(\frac{\theta}{2}) d\theta \) which simplifies to \( \int_{0}^{\pi} 1 d\theta \), giving the area as \( \pi \).

Key Concepts

Integration by Substitution

Integration by Substitution is a powerful technique in calculus for evaluating complex integrals. It is akin to solving a puzzle by changing the pieces involved, making them fit together more easily. This method involves substituting parts of the integral with a new variable, simplifying the expression.

For instance, in the solution to the integral \( \int_{0}^{1} \sqrt{\frac{1-x}{x}} \, dx \),we introduced a new variable.

• Let \( x = \frac{1}{1 + y^2} \), implying that \( y = \sqrt{\frac{1-x}{x}} \)
• Differentiating gives \( dx = -\frac{2y}{(1+y^2)^2} \, dy \)

This substitution simplifies the integration process significantly. By transforming the variables, we manage to express the integral in a simpler form that is easier to solve. Understanding how to choose effective substitutions, and performing the corresponding transformations, is invaluable when tackling complex integrals.

Change of Variables in Integration

Change of variables, often used alongside Integration by Substitution, transforms the limits and the function being integrated. This method redefines the region where integration occurs. It can convert an otherwise difficult integral into one that is straightforward to evaluate.

In the exercise, when solving part (b) with integral \( \int_{-1}^{1} \sqrt{\frac{1+x}{1-x}} \, dx \):

• We used \( x = \cos(\theta) \) as a substitution.
• Because of this transformation, the differential becomes \( dx = -\sin(\theta) \, d\theta \).

The integral's new limits change from the original \( x \, \text{from} \, -1 \, \text{to} \, 1 \) to \( \theta \, \text{from} \, 0 \, \text{to} \, \pi \). Such variable transformations are crucial in making integrals more manageable by choosing coordinates or domains that make the problem simpler.

Area Under a Curve

The concept of the area under a curve connects geometry with calculus. An integral can be visualized as the accumulated area under a curve over an interval on the x-axis. This is essentially the geometric representation of integration.

When addressing the integral \( \int_{0}^{1} \sqrt{\frac{1-x}{x}} \, dx \):

• Think of it as calculating the area under the curve formed by \( y = \sqrt{\frac{1-x}{x}} \) from \( x = 0 \) to \( x = 1 \).

Similarly, for \( \int_{-1}^{1} \sqrt{\frac{1+x}{1-x}} \, dx \):

• This represents finding the area under \( y = \sqrt{\frac{1+x}{1-x}} \) between \( x = -1 \) and \( x = 1 \).

By understanding the integral as a measure of area, it becomes easier to apply and interpret these processes beyond just the numerical computation.

Calculus Problem Solving

Solving calculus problems requires a blend of theoretical knowledge and practical techniques. It often involves a step-by-step approach to break down complex problems into manageable parts. Each step in evaluating an integral is crucial in leading towards the solution.

The problem of evaluating integrals, such as given in this exercise, can be tackled with a methodical approach:

• Identify the integrand and the region of integration.
• Decide if any substitutions or transformations can simplify the integral.
• Perform the calculations faithfully, making sure to transform limits appropriately.
• Interpret the result in the context of the problem, like assessing calculated areas.

By following these steps, one not only arrives at the right numerical answer but also builds a deeper understanding of calculus concepts. This structured approach is integral in handling everything from classroom exercises to more advanced applications in physics and engineering.