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Chapter 7: Problem 40

Make the \(u\) -substitution and evaluate the resulting definite integral. $$ \int_{0}^{+\infty} \frac{e^{-x}}{\sqrt{1-e^{-2 x}}} d x ; u=e^{-x} $$

Short Answer

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

Step by step solution

01

Apply the Substitution

We are given to use the substitution \( u = e^{-x} \). Therefore, differentiate \( u \) with respect to \( x \) to find the differential \( du \): \[\frac{du}{dx} = -e^{-x} = -u\]Thus, the differential \( dx \) becomes: \[dx = -\frac{du}{u}\]
02

Change the Limits of Integration

Before substituting, we need to change the limits of integration according to the substitution \( u = e^{-x} \):- When \( x = 0 \), \( u = e^{0} = 1 \).- When \( x = \infty \), \( u = e^{- ext{infinity}} = 0 \).So the new limits of integration are from \( 1 \) to \( 0 \).
03

Substitute in the Integral

Substitute \( u = e^{-x} \) and \( dx = -\frac{du}{u} \) into the integral:\[\int_{0}^{\infty} \frac{e^{-x}}{\sqrt{1-e^{-2x}}} \, dx = \int_{1}^{0} \frac{u}{\sqrt{1-u^2}} \left(-\frac{du}{u}\right)\]This simplifies to:\[- \int_{1}^{0} \frac{1}{\sqrt{1-u^2}} \, du\]Now, reverse the limits of integration to remove the negative sign:\[\int_{0}^{1} \frac{1}{\sqrt{1-u^2}} \, du\]
04

Evaluate the Integral

The integral \( \int \frac{1}{\sqrt{1-u^2}} \, du \) is a standard integral, representing the arcsine function:\[\int \frac{1}{\sqrt{1-u^2}} \, du = \arcsin(u) + C\]Thus, evaluate the definite integral from \( 0 \) to \( 1 \):\[\left[ \arcsin(u) \right]_{0}^{1} = \arcsin(1) - \arcsin(0)\]Since \( \arcsin(1) = \frac{\pi}{2} \) and \( \arcsin(0) = 0 \), the result is:\[\frac{\pi}{2} - 0 = \frac{\pi}{2}\]
05

Conclusion

The definite integral evaluates to \( \frac{\pi}{2} \), which is the final result after the \( u \)-substitution and integration steps.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

definite integral
A definite integral is a mathematical concept that calculates the area under a curve, defined by a specific function, between two specified limits or bounds. It's not just about accumulating values but also indicates net change in a quantity described by the function over a certain interval. The integral itself is defined as:
  • The symbol \( \int \) represents integration.
  • The limits, usually noted as lower and upper limits, provide specific boundaries for the integration.
  • The function inside the integral sign is the integrand, which describes the curve we are finding the area under.
  • Finally, the last piece is \( dx \), which signifies the variable of integration.
When evaluating a definite integral, a geometric interpretation often involves calculating the exact area under a curve between two points on a graph. In this exercise, the integration was performed by transforming the function using a suitable substitution which simplified solving the integral.
change of variables
The change of variables technique, also known as u-substitution, is a powerful method in calculus used to simplify integrals. This method involves replacing a complex expression within an integral with a simpler variable. Here's how it works:
  • Select a new variable \( u \) that replaces a part of the original integrand with the aim of simplifying it.
  • Determine \( du \) by differentiating \( u \) with respect to \( x \), and then express \( dx \) in terms of \( du \) and \( u \).
  • Convert the limits of integration to match the new variable \( u \). Transition them from original \( x \) terms to \( u \) terms, considering the initial given boundaries.
  • Substitute all elements of the integral with the new \( u \) expressions and solve the simplified integral.
In this specific example, the substitution \( u = e^{-x} \) was chosen. The change of variables made the problem simpler by transforming the integrand into a form that is easier to integrate.
arcsine function
The arcsine function, often denoted as \( \arcsin \), is the inverse of the sine function on the interval \(-1, 1\). Its primary role in calculus and this exercise is to provide the antiderivative of \( \frac{1}{\sqrt{1-u^2}} \), which is a common form encountered in integration involving trigonometric functions. Here鈥檚 what you should know:
  • \( \arcsin(u) \) returns the angle whose sine is \( u \).
  • The typical range of the arcsine function is \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\), limiting the results to the first and fourth quadrants.
  • When calculating definite integrals using the arcsine, remember that the evaluated result represents an angle.
In the original problem, after substitution and simplification, the given integral became a standard form whose solution is expressed using the arcsine function. This transformation greatly simplified the calculation of the defined area (integral), reaching a final answer efficiently as \( \frac{\pi}{2} \).

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Most popular questions from this chapter

What is "improper" about a definite integral over an interval on which the integrand has an infinite discontinuity? Explain why Definition \(5.5 .1\) for \(\int_{a}^{b} f(x) d x\) fails if the graph of \(f\) has a vertical asymptote at \(x=a\). Discuss a strategy for assigning a value to \(\int_{a}^{b} f(x) d x\) in this circumstance.

Transform the given improper integral into a proper integral by making the stated \(u\) -substitution; then approximate the proper integral by Simpson's rule with \(n=10\) subdivisions. Round your answer to three decimal places. $$ \int_{0}^{1} \frac{\sin x}{\sqrt{1-x}} d x ; u=\sqrt{1-x} $$

Let \(R\) be the region to the right of \(x=1\) that is bounded by the \(x\) -axis and the curve \(y=1 / x .\) When this region is revolved about the \(x\) -axis it generates a solid whose surface is known as Gabriel's Horn (for reasons that should be clear from the accompanying figure). Show that the solid has a finite volume but its surface has an infinite area. [Note: It has been suggested that if one could saturate the interior of the solid with paint and allow it to seep through to the surface, then one could paint an infinite surface with a finite amount of paint! What do you think?]

(a) Make \(u\) -substitution (5) to convert the integrand to a rational function of \(u\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral (no substitution), and then confirm that the result is equivalent to that in part (a). $$ \int \frac{d \theta}{1-\cos \theta} $$

Suppose that the region between the \(x\) -axis and the curve \(y=e^{-x}\) for \(x \geq 0\) is revolved about the \(x\) -axis. (a) Find the volume of the solid that is generated. (b) Find the surface area of the solid.
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