/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Find the general antiderivative ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the general antiderivative of the given function. $$ f(x)=5 e^{3 x}-\sec ^{2}(x-3) $$

Short Answer

Expert verified
The general antiderivative is \( F(x) = \frac{5}{3}e^{3x} - \tan(x-3) + C \).

Step by step solution

01

Identifying the Components

The function given is \( f(x) = 5e^{3x} - \sec^2(x-3) \). We need to find the antiderivative of each term separately.
02

Antiderivative of the Exponential Function

For the term \( 5e^{3x} \), we use the fact that the antiderivative of \( e^{ax} \) is \( \frac{1}{a}e^{ax} \). Thus, the antiderivative of \( 5e^{3x} \) is \( \frac{5}{3}e^{3x} \).
03

Antiderivative of the Hyperbolic Secant Squared Function

The antiderivative of \(-\sec^2(u)\) is \(-\tan(u)\). Here, \(u = x-3\). Using substitution, the antiderivative becomes \(-\tan(x-3)\).
04

Combining Antiderivatives

Combine the antiderivatives from the previous steps and add the constant of integration. The general antiderivative is \( F(x) = \frac{5}{3}e^{3x} - \tan(x-3) + C \), where \( C \) is an arbitrary constant.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Exponential Function
An exponential function is one of the most fascinating mathematical functions. The general form is \( e^{ax} \), where \( e \) is the base of the natural logarithm, approximately 2.718. This base is special in mathematics because it leads to a unique derivative and antiderivative identity: the function remains the same, up to a scalar multiple.

In the context of our exercise, we have \( 5e^{3x} \). Here, the exponent has been scaled by a constant factor of 3, referred to as the rate of the exponential growth or decay. The antiderivative of such a function is nothing but a scaled version of the same function. Specifically, integrating \( e^{ax} \) results in \( \frac{1}{a}e^{ax} \), reflecting the unique property of exponential functions where the derivative or antiderivative is proportional to the original function.

Thus, the antiderivative of \( 5e^{3x} \) becomes \( \frac{5}{3}e^{3x} \). This simplicity and elegance make exponential functions pivotal in both mathematics and applied sciences. Understanding this concept can also aid in solving more complex calculus problems.
Integration Technique
Integration is essentially the reverse process of differentiation in calculus, and it's used to calculate antiderivatives. The primary goal is to determine a function whose derivative equals the given function. There are several integration techniques available, each tailored for handling specific types of functions effectively.

For exponential functions, as mentioned earlier, the integration process is straightforward because of their self-derivative property. However, the integration of trigonometric functions like \(-\sec^2(u)\) requires recognition of standard derivative forms. The derivative of \(-\tan(u)\) results in \(-\sec^2(u)\), making the antiderivative \(-\tan(u)\). In this exercise, we dealt with a linear expression within the secant function, requiring a change of variable for integration: substituting \( u = x - 3 \). This substitution simplifies the integration process.

Using the constants and understanding these standard forms facilitates easier computation of antiderivatives in problems like \( f(x) = 5e^{3x} - \sec^2(x-3) \). Recognizing these patterns not only simplifies current challenges but also builds the foundation for more advanced calculus topics.
Secant Function
The secant function, noted as \( \sec(x) \), is the reciprocal of the cosine function, making it a crucial component in trigonometry. It is defined as \( \sec(x) = \frac{1}{\cos(x)} \). When squaring the secant function, such as \( \sec^2(x) \), it represents one of the standard trigonometric identities encountered in calculus involving derivatives and antiderivatives.

In our exercise, the term \(-\sec^2(x-3)\) was integrated. The integral of \(-\sec^2(u)\), where \( u \) is a function of \( x \), leads to \(-\tan(u)\). Since the derivative \( \frac{d}{dx}(\tan(x)) = \sec^2(x) \), it follows that the antiderivative mirrors the tangent's differentiation. Here, substituting \( u = x-3 \) allowed us to directly apply this relationship, arriving at \(-\tan(x-3)\) as part of our antiderivative solution.

This illustrates the importance of understanding and using trigonometric identities, and the secant function's role in them, to solve calculus problems efficiently. Mastery of these identities is crucial for higher-level mathematics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve by rewriting the differential equation as an equation for \(\frac{d x}{d y}\) : $$ \frac{d y}{d x}=1-y, \text { for } x \geq 0 \text { with } y(0)=0 $$

The Ricker model was introduced by Ricker (1954) as an alternative to the discrete logistic equation to describe the density-dependent growth of a population. Under the Ricker model the population \(N_{t}\) sampled at discrete times \(t=0,1,2, \ldots\) is modeled by a recurrence equation $$N_{t+1}=R_{0} N_{t} \exp \left(-a N_{t}\right)$$ where \(R_{0}\) and \(a\) are positive constants that will vary between different species and between different habitats. (a) Explain why you would expect \(R_{0}>1\) (Hint: consider the population growth when \(N_{t}\) is very small.) (b) Show that the recursion relation has two equilibria, a trivial equilibrium (that is, \(N=0\) ) and another equilibrium, which you should find. (c) Show that if \(R_{0}>1\) then use the stability criterion for equilibria to show that the trivial equilibrium point is unstable. (d) Use the stability criterion for equilibria to show that the nontrivial equilibrium point is stable if \(0<\ln R_{0}<2\). (e) If \(R_{0}>1\) then \(\ln R_{0}>0\), so most populations will meet the first inequality condition. What happens if \(\ln R_{0}>2 ?\) Let's try some explicit values: \(R_{0}=10, a=1, N_{0}=1 .\) Calculate the first ten terms of the sequence, and describe in words how the sequence behaves.

Hill's equation for the oxygen saturation of blood states that the level of oxygen saturation (fraction of hemoglobin molecules that are bound to oxygen) in blood can be represented by a function: $$ f(P)=\frac{P^{n}}{P^{n}+30^{n}} $$ where \(P\) is the oxygen concentration around the blood \((P \geq 0)\) and \(n\) is a parameter that varies between different species. (a) Assume that \(n=1\). Show that \(f(P)\) is an increasing function of \(P\) and that \(f(P) \rightarrow 1\) as \(P \rightarrow \infty\). (b) Assuming that \(n=1\) show that \(f(P)\) has no inflection points. Is it concave up or concave down everywhere? (c) Knowing that \(f(P)\) has no inflection points, could you deduce which way the curve bends (whether it is concave up or concave down) without calculating \(f^{\prime \prime}(P) ?\) (d) For most mammals \(n\) is close to 3. Assuming that \(n=3\) show that \(f(P)\) is an increasing function of \(P\) and that \(f(P) \rightarrow 1\) as \(P \rightarrow \infty\) (e) Assuming that \(n=3\), show that \(f(P)\) has an inflection point, and that it goes from concave up to concave down at this inflection point. (f) Using a graphing calculator plot \(f(P)\) for \(n=1\) and \(n=3\). How do the two curves look different?

Consider the following discrete logistic model for the change in the size of a population over time: $$N_{t+1}=R_{0} N_{t}-\frac{1}{100} N_{t}^{2}$$ for \(t=0,1,2, \ldots\) (a) Find all equilibria when \(R_{0}=2.5\) and calculate which (if any) are stable. (b) Calculate the first ten terms of the sequence when \(N_{0}=10\) and describe what you see.

Suppose that a drug is eliminated so slowly from the blood that its elimination kinetics can be essentially ignored. Then according to Section \(5.9\) the total amount of drug in the blood is given by a differential equation: $$ \frac{d M}{d t}=A(t) $$ where \(A(t)\) is the rate of absorption. We will show in Chapter 8 that if the drug is absorbed into the blood from a pill in the patient's gut, then \(A(t)\) is given by a function $$ A(t)=C e^{-k t} $$ where \(C>0\) and \(k>0\) are constants that depend on the type of the drug being administered. Assume that at \(t=0\) there is no drug present in the patient's blood (i.e., \(M(0)=0\) ). Solve this initial value problem, and, using the methods from Section \(5.6\), sketch the graph of \(M(t)\) against \(t\).

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.