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Chapter 7: Problem 7

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

Short Answer

Expert verified
\( \int e^x \sinh(e^x) \, dx = \cosh(e^x) + C \).

Step by step solution

01

Identify the Inner Function

Notice that the integrand is a composite function involving the exponential function and the hyperbolic sine function. The argument of the hyperbolic sine function, \( e^x \), suggests a suitable \( u \)-substitution. Let's set \( u = e^x \), making its derivative, \( du = e^x dx \). Therefore, \( dx = \frac{du}{e^x} \), but because \( e^x = u \), we have \( dx = \frac{du}{u} \).
02

Rewrite the Integral Using Substitution

Substitute \( u = e^x \) into the integral to obtain:\[ \int e^x \sinh(e^x) \, dx = \int u \sinh(u) \cdot \frac{du}{u} = \int \sinh(u) \, du \] Here, the \( e^x \) terms cancel out since \( u = e^x \).
03

Integrate the Simplified Function

Now, we focus on the integral \( \int \sinh(u) \, du \). Recall the integration rule for hyperbolic sine:\[ \int \sinh(u) \, du = \cosh(u) + C \]Thus, \( \int \sinh(u) \, du = \cosh(u) + C \).
04

Back-Substitution

Re-substitute \( u = e^x \) back into the expression:\[ \cosh(u) + C = \cosh(e^x) + C \]Therefore, the integral evaluates to \( \cosh(e^x) + C \), where \( C \) is the constant of integration.

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Key Concepts

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

u-substitution
U-substitution is a handy technique in calculus to simplify and solve an integral. It's like renaming a complex expression to something simpler, making integration easier. In essence, we identify a part of the integrand (the function being integrated) that can be replaced by a single variable, usually noted by "u".

This technique is particularly useful when dealing with composite functions, that is, functions within functions.
  • First, we identify a complicated expression and set it equal to "u". In our exercise, this is the exponential part, so we set \( u = e^x \).
  • The next step is to differentiate this expression with respect to \( x \), obtaining \( du = e^x \, dx \), which rearranges to \( dx = \frac{du}{e^x} \).
  • By substituting these into the integral, we simplify the process. Here, the integrand with \( e^x \) cancels out properly.
With substitution, the seemingly complex integral becomes much more straightforward, allowing integration rules to be applied easily. This makes u-substitution an indispensable tool for tackling integrals featuring composite functions.
hyperbolic functions
Hyperbolic functions are analogs of trigonometric functions but for a hyperbola, instead of a circle. They have several important applications in calculus and beyond, much like their circular counterparts.
  • The hyperbolic sine function, \( \sinh(x) \), is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).
  • Hyperbolic functions solve equations related to hyperbolic relationships, which are common in physics and engineering.
In terms of integration, hyperbolic functions obey similar rules to trigonometric functions.

When we integrate the hyperbolic sine, we apply the rule: \( \int \sinh(u) \, du = \cosh(u) + C \), where \( \cosh(u) = \frac{e^u + e^{-u}}{2} \). This part of our exercise shows just how useful these functions are when simplifying and evaluating integrals that involve composite functions.
exponential function
The exponential function, \( e^x \), is one of the most significant functions in calculus due to its unique properties and its ability to model growth and decay processes in nature.
  • It is defined as \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \), an infinite series that converges for any real number \( x \).
  • The derivative and integral of \( e^x \) are both \( e^x \), which is a rare mathematical phenomenon making it very powerful in calculus.
In our problem, the exponential function is both part of the inner function and the multiplier in the integrand, arising from the function \( \sinh(e^x) \). This contributes to the choice of substitution and simplification process.

Ultimately, the characteristic of the exponential function is crucial not only for transforming and simplifying expressions but also for connecting various types of functions, such as hyperbolic functions in this instance.

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Most popular questions from this chapter

Determine whether the statement is true or false. Explain your answer. \(\int_{1}^{2} \frac{1}{x(x-3)} d x\) is an improper integral.

Suppose that during a period \(t_{0} \leq t \leq t_{1}\) years, a company has a continuous income stream at a rate of \(I(t)\) dollars per year at time \(t\) and that this income is invested at an annual rate of \(r \%,\) compounded continuously. The value (in dollars) of this income stream at the end of the time period \(t_{0} \leq t \leq t_{1},\) called the stream's future value, can be calculated using $$F V=\int_{t_{0}}^{t_{1}} I(t) e^{r\left(t_{1}-t\right)} d t$$ The present value (in dollars) of the income stream is given by $$P V=\int_{t_{0}}^{t_{1}} I(t) e^{-r\left(t-t_{0}\right)} d t$$ The present value is the amount that, if put in the bank at time \(t=t_{0}\) at \(r \%\) compounded continuously, with no additional deposits, would result in a balance of \(F V\) dollars at time \(t=t_{1}\) That is, $$F V=P V \times e^{r\left(t_{1}-t_{0}\right)}$$ In each exercise, (a) find the future value \(F V\) for the given income stream \(I(t)\) and interest rate \(r\) and time period \(t_{0} \leq t \leq t_{1}\) (b) find the present value \(P V\) of the income stream over the time period; and (c) verify that \(F V\) and \(P V\) satisfy the relationship given above. $$I(t)=20,000-200 t^{2} ; r=5 \% ; 2 \leq t \leq 10$$

What is "improper" about an integral over an infinite interval? Explain why Definition 5.5 .1 for \(\int_{a}^{b} f(x) d x\) fails for \(\int_{a}^{+\infty} f(x) d x .\) Discuss a strategy for assigning a value to \(\int_{a}^{+\infty} f(x) d x\).

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.

Use any method to find the volume of the solid generated when the region enclosed by the curves is revolved about the \(y\) -axis. $$y=\sqrt{x-4}, y=0, x=8$$
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