Problem 19 - University Calculus: Early Transcendentals

Question

Evaluate the integrals in Exercises \(7-20\). $$\int_{0}^{\pi / 4} \int_{0}^{\ln \sec v} \int_{-\infty}^{2 t} e^{x} d x d t d v \quad(t v x-\text { space })$$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{1}{2} - \frac{\pi}{8}\).

Step 1: Understanding the Integral Setup

The given integral is a triple integral and must be evaluated in the order of \( dx \, dt \, dv \). The boundaries are: \(-\infty \) to \(2t\) for \(x\), 0 to \(\ln \sec v\) for \(t\), and 0 to \(\pi/4\) for \(v\). The integrand is \( e^x \).

Step 2: Evaluate the Innermost Integral

First, we evaluate the innermost integral with respect to \(x\). It is given by \( \int_{-\infty}^{2t} e^x \, dx \). The antiderivative of \( e^x \) is \( e^x \), and due to the exponential decay at negative infinity, we compute:\[ \int_{-\infty}^{2t} e^x \, dx = \left[e^x\right]_{-\infty}^{2t} = e^{2t} - 0 = e^{2t} \]

Step 3: Evaluate the Second Integral

Substitute the result from Step 2 into the second integral. We need to evaluate \( \int_{0}^{\ln \sec v} e^{2t} \, dt \). The antiderivative of \( e^{2t} \) is \( \frac{1}{2} e^{2t} \). Evaluating the definite integral gives:\[ \int_{0}^{\ln \sec v} e^{2t} \, dt = \left[ \frac{1}{2} e^{2t} \right]_{0}^{\ln \sec v} = \frac{1}{2} e^{2(\ln \sec v)} - \frac{1}{2} e^0 = \frac{1}{2} (\sec^2 v - 1) \]

Step 4: Evaluate the Outer Integral

Finally, evaluate the outermost integral with respect to \(v\). Substitute the result from Step 3 into the integral:\[ \int_{0}^{\pi/4} \frac{1}{2} (\sec^2 v - 1) \, dv \]This can be split into two integrals:\[ = \frac{1}{2} \int_{0}^{\pi/4} \sec^2 v \, dv - \frac{1}{2} \int_{0}^{\pi/4} \, dv \]The antiderivative of \( \sec^2 v \) is \( \tan v \), and for constant 1, it is \( v \). Evaluate each:\[ \frac{1}{2} [\tan v]_0^{\pi/4} = \frac{1}{2} (1 - 0) = \frac{1}{2} \]\[ \frac{1}{2} [v]_0^{\pi/4} = \frac{1}{2} (\frac{\pi}{4} - 0) = \frac{\pi}{8} \]Thus, the combined result is:\[ \frac{1}{2} - \frac{\pi}{8} \]

Step 5: Final Computation

Combine the results from step 4 to get the final value of the entire triple integral:\[ \frac{1}{2} - \frac{\pi}{8} \]

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, providing a way to calculate the net area under a curve over a specific interval. This is particularly useful when dealing with functions where the range of values needs to be considered in the context of accumulation or total effect. In the given exercise, the use of definite integrals comes into play when evaluating each layer of the triple integral.

• The innermost integral spans from \(-\infty\) to \(2t\) over the variable \(x\), capturing the exponential decay towards negative infinity.
• The second integral, from 0 to \(\ln \sec v\), considers the effect of \(t\) over this interval.
• The outermost integral, from 0 to \(\pi/4\), integrates with respect to \(v\), finalizing the computation over the specified bounds.

Each step relies on solving definite integrals, systematically reducing the dimensions of the triple integral until a single numerical value is achieved.

Integration by Parts

Integration by parts is a technique used to transform the integral of a product of functions into a potentially more manageable form. Though not explicitly used in this exercise, understanding this method aids in grasping how complex integrals are often tackled.In essence, integration by parts is derived from the product rule for differentiation and is expressed by the formula:\[ \int u \, dv = uv - \int v \, du \]Where:

• \(u\) and \(dv\) are parts of the original function.
• \(du\) and \(v\) are derivatives and integrals of the chosen components, respectively.

This principle would be particularly useful if we encountered an integral involving a distinct product of functions, where one is easily integrable, like an exponential or trigonometric function, paired with a polynomial.

Exponential Functions

Exponential functions play a pivotal role in the exercise, primarily through the integrand \(e^x\). These functions are characterized by their constant proportionality, as they grow or decay at rates proportional to their current value. They appear frequently in calculus because of their unique property where the rate of change is proportional to the function's value.In this problem:

• The innermost integral exploits the property that the derivative and antiderivative of \(e^x\) is \(e^x\) itself.
• Thus, the integral of \(e^x\) from \(-\infty\) to \(2t\) relies on the effect of exponential growth towards this bound, producing \(e^{2t}\).

This characteristic simplifies the computation significantly, especially when combined with bounds that include infinity, where the function’s decay neutralizes potentially divergent terms.

Calculus

Calculus, as a comprehensive mathematical field, consists of differential and integral calculus. In this exercise, the calculus principles of integration are prominently showcased. By breaking down the problem into smaller sections, each integration task builds on previously computed results.Key aspects of calculus evident in this problem include:

• The fundamental theorem of calculus, linking definite integrals and antiderivatives, facilitates the computation at each step.
• The limit and continuity principles help manage infinite bounds and apply appropriate convergence criteria.
• Separate integration of functions like \(\sec^2 v\) and constants reflect differential calculus when deriving antiderivatives and integral calculus when evaluating them over defined intervals.

Altogether, these foundational calculus concepts provide the toolkit required to systematically tackle and solve multi-dimensional integrals like this one.