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Chapter 7: Problem 30

Evaluate the integrals by making appropriate \(u\) -substitutions and applying the formulas reviewed in this section. $$ \int 2^{\pi x} d x $$

Short Answer

Expert verified
\( \int 2^{\pi x} \, dx = \frac{2^{\pi x}}{\pi \ln(2)} + C.\)

Step by step solution

01

Identify a Suitable Substitution

To evaluate the integral \( \int 2^{\pi x} \, dx \), we start by identifying a substitution that simplifies the expression. Notice that the power \( \pi x \) in the exponent suggests using \( u = \pi x \). Thus, we have \( du = \pi \, dx \) and therefore \( dx = \frac{1}{\pi} \, du \).
02

Rewrite the Integral with Substitution

Substituting \( u = \pi x \) and \( dx = \frac{1}{\pi} \, du \) into the integral gives us:\[\int 2^{u} \cdot \frac{1}{\pi} \, du = \frac{1}{\pi} \int 2^{u} \, du.\]
03

Integrate Using the Exponential Formula

The integral \( \int 2^{u} \, du \) can be solved using the formula for integrating exponential functions: \( \int a^{u} \, du = \frac{a^{u}}{\ln(a)} + C \), where \( a = 2 \). Substituting into the formula, we have:\[\frac{1}{\pi} \int 2^{u} \, du = \frac{1}{\pi} \left( \frac{2^{u}}{\ln(2)} \right) + C.\]
04

Back-Substitute to Original Variable

Replace \( u \) with \( \pi x \) to express the solution in terms of the original variable:\[= \frac{1}{\pi \ln(2)} \cdot 2^{\pi x} + C.\]

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Key Concepts

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

Integrals
When we talk about integrals in mathematics, we're delving into a concept that is crucial for calculus. An integral essentially represents the area under a curve, or more technically, it provides the accumulation of quantities, such as area, volume, or other continuous quantities.
Definite integrals, denoted as \( \int_{a}^{b} f(x) \, dx \), calculate the area under a curve from point \(a\) to \(b\). In contrast, indefinite integrals, like \( \int f(x) \, dx \), provide a family of functions as their solution, representing an antiderivative.
Integrals are fundamental in solving many different problems in physics and engineering where they help calculate work done, centers of mass, and volumes.鈥� This concept is primarily grounded in the mystery of how things accumulate, be it time, matter, or energy. Learning to approach integrals as tools helps us manage and predict physical phenomena.
Exponential Functions
Exponential functions are a special category of mathematical functions where the variable appears in the exponent. They have the general form \( f(x) = a^{x} \), where \( a \) is a constant greater than zero.
These functions exhibit rapid growth (or decay) and are important in describing phenomena in fields like population growth, radioactive decay, and interest calculations.
When integrating exponential functions, things become quite intriguing. The nature of exponential growth makes them quite predictable and perfectly suited for calculus operations. For instance, integrating \( a^{x} \) requires using the formula: \( \int a^{x} \, dx = \frac{a^{x}}{\ln(a)} + C \). Thus, these functions integrate into simpler expressions, yet remain exponential in nature.
Understanding how exponential functions behave and how they can be manipulated through integration is essential in both pure and applied mathematics. Their predictability in growth patterns underpins many scientific and financial models.
Integration Techniques
Integration techniques are the methods used to solve integrals and make complex integrals more manageable. One popular technique is **u-substitution**. This technique involves changing variables to simplify the integral into a basic, recognizable form.
To use u-substitution effectively, one must:
  • Identify a part of the integral that can be substituted, typically setting \( u = g(x) \) where \( g(x) \) is a function whose derivative \( g'(x) \) is present in the integrand.

  • Change the differential \( dx \) by calculating the derivative \( du = g'(x) \, dx \), providing \( dx = \frac{1}{g'(x)} \, du \).

  • Rewrite the integral using \( u \) in place of \( g(x) \) and \( dx \). This transforms it into an easier form to integrate.

  • Finally, after integrating, transform back to the original variable using the original substitution equation.

Mastering this technique is invaluable, especially when tackling integrals involving exponential functions or trigonometric identities. It allows the simplification of an otherwise unapproachable integral, streamlining the integration process considerably, and is a fundamental skill for anyone studying calculus.

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Most popular questions from this chapter

(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). $$ \int \frac{d \theta}{1-\cos \theta} $$

(a) Make an appropriate \(u\) -substitution of the form \(u=x^{1 / n}\) or \(u=(x+a)^{1 / n}\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$ \int \frac{x}{\sqrt{x+1}} d x $$

Medication can be administered to a patient using a variety of methods. For a given method, let \(c(t)\) denote the concentration of medication in the patient's bloodstream (measured in \(\mathrm{mg} / \mathrm{L}) t\) hours after the dose is given. The area under the curve \(c=c(t)\) over the time interval \([0,+\infty)\) indicates the "availability" of the medication for the patient's body. Determine which method provides the greater availability. Method \(1: c_{1}(t)=5\left(e^{-0.2 t}-e^{-t}\right)\) Method 2: \(c_{2}(t)=4\left(e^{-0.2 t}-e^{-3 t}\right)\)

In each part, confirm the result with a CAS. (a) \(\int_{0}^{+\infty} \frac{\sin x}{\sqrt{x}} d x=\sqrt{\frac{\pi}{2}}\) (b) \(\int_{-\infty}^{+\infty} e^{-x^{2}} d x=\sqrt{\pi}\) (c) \(\int_{0}^{1} \frac{\ln x}{1+x} d x=-\frac{\pi^{2}}{12}\)

Determine whether the statement is true or false. Explain your answer. The Simpson's rule approximation \(S_{50}\) for \(\int_{a}^{b} f(x) d x\) is a weighted average of the approximations \(M_{50}\) and \(T_{50}\), where \(M_{50}\) is given twice the weight of \(T_{50}\) in the average.
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