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Chapter 6: Problem 23

Evaluate the following integrals. \(\int \frac{e^{\sqrt{x}}}{\sqrt{x}} d x\)

Short Answer

Expert verified
Short Answer: Using substitution method, the integral \(\int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx\) can be evaluated as \(2e^{\sqrt{x}}+C\).

Step by step solution

01

Choose a substitution

Let \(u=\sqrt{x}\). Then \(u^2=x\). Now differentiate both sides with respect to \(x\) to find the necessary replacement for \(dx\).
02

Differentiate the substitution

Differentiate \(u^2=x\) with respect to \(x\): \(2u\frac{du}{dx} = 1 \Rightarrow \frac{du}{dx} = \frac{1}{2u}.\) Now solve for \(dx\): \(dx = 2u\,du.\)
03

Rewrite the integral in terms of \(u\)

Using our substitutions, we rewrite the integral as: \(\int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx = \int \frac{e^u}{u}(2u\,du) = 2\int e^u du\)
04

Solve the integral in terms of \(u\)

The integral \(\int e^u du\) is relatively straightforward to compute: \(2\int e^u du = 2e^u+C\), where \(C\) represents the constant of integration.
05

Return to the original variable \(x\)

Substitute \(\sqrt{x}\) back in for \(u\): \(2e^u+C = 2e^{\sqrt{x}}+C\)
06

Write the final answer

So, the final answer is : \(\int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx = 2e^{\sqrt{x}}+C.\)

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Key Concepts

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

Definite Integrals
Definite integrals allow us to find the accumulated quantity, like the area under a curve, between two points. They emerge from the concept of integration, which is essentially the sum of infinitesimal quantities. In the original exercise, however, we dealt with an indefinite integral since we did not specify limits of integration.
Unlike definite integrals, which include limits and result in a specific numerical value, indefinite integrals do not have limits and include a constant of integration, denoted as \(C\).
  • To solve definite integrals, we use the Fundamental Theorem of Calculus, which connects differentiation and integration.
  • This theorem states that, essentially, integration could be considered as "inverting" differentiation.
  • Definite integrals can represent various physical quantities such as areas, volumes, and even net change.
Understanding definite vs. indefinite integrals is crucial for solving many problems in calculus, such as those involving areas under curves and accumulative quantities.
Exponential Functions
Exponential functions are mathematical functions of the form \(f(x) = a^x\), where \(a\) is a constant and \(x\) is the exponent. They are wildly important in calculus and have distinctive properties that make them unique.
In the context of calculus, particularly integrals, they are invaluable because:
  • They have straightforward derivatives and integrals. For example, the derivative of \(e^x\) is \(e^x\) itself.
  • This feature makes calculations involving them simpler, as demonstrated in the original exercise. The integration of \(e^u\) with respect to \(u\) yields \(e^u+C\), drastically simplifying the process.
  • Exponential growth and decay models in real-world scenarios are often based on exponential functions, explaining phenomena like population growth or radioactive decay.
By harnessing these properties, exponential functions remain a vital component of continuous mathematical modeling and analytical solving.
Differential Calculus
Differential calculus focuses on the concept of a derivative, which describes how a function changes as its input changes. It's all about rates of change and slopes of curves.
Using substitution in the context of integration is an application of differential calculus concepts. In our example:
  • We employed substitution \(u = \sqrt{x}\) to simplify the function. This step involves recognizing patterns and strategically applying derivatives.
  • By differentiating \(x = u^2\), we obtain \(\frac{dx}{du} = 2u\), allowing us to replace \(dx\) with \(2u\,du\) in the integral calculation.
  • Mastery of differential calculus is essential for simplifying and solving integrals, as it bridges different mathematical expressions into manageable forms.
Thus, differential calculus isn't just about differentiating functions; it's a versatile toolset for transforming and solving complex integral problems.

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Most popular questions from this chapter

Carry out the following steps to derive the formula \(\int \operatorname{csch} x d x=\ln |\tanh (x / 2)|+C(\text { Theorem } 6.9)\) a. Change variables with the substitution \(u=x / 2\) to show that $$\int \operatorname{csch} x d x=\int \frac{2 d u}{\sinh 2 u}.$$ b. Use the identity for sinh \(2 u\) to show that \(\frac{2}{\sinh 2 u}=\frac{\operatorname{sech}^{2} u}{\tanh u}.\) c. Change variables again to determine \(\int \frac{\operatorname{sech}^{2} u}{\tanh u} d u\) and then express your answer in terms of \(x.\)

Determine whether the following statements are true and give an explanation or counterexample. a. When using the shell method, the axis of the cylindrical shells is parallel to the axis of revolution. b. If a region is revolved about the \(y\) -axis, then the shell method must be used. c. If a region is revolved about the \(x\) -axis, then in principle, it is possible to use the disk/washer method and integrate with respect to \(x\) or the shell method and integrate with respect to \(y\)

A rigid body with a mass of 2 kg moves along a line due to a force that produces a position function \(x(t)=4 t^{2},\) where \(x\) is measured in meters and \(t\) is measured in seconds. Find the work done during the first \(5 \mathrm{s}\) in two ways. a. Note that \(x^{\prime \prime}(t)=8 ;\) then use Newton's second law \(\left(F=m a=m x^{\prime \prime}(t)\right)\) to evaluate the work integral \(W=\int_{x_{0}}^{x_{1}} F(x) d x,\) where \(x_{0}\) and \(x_{f}\) are the initial and final positions, respectively. b. Change variables in the work integral and integrate with respect to \(t .\) Be sure your answer agrees with part (a).

Suppose a cylindrical glass with a diameter of \(\frac{1}{12} \mathrm{m}\) and a height of \(\frac{1}{10} \mathrm{m}\) is filled to the brim with a 400-Cal milkshake. If you have a straw that is 1.1 m long (so the top of the straw is \(1 \mathrm{m}\) above the top of the glass), do you burn off all the calories in the milkshake in drinking it? Assume that the density of the milkshake is \(1 \mathrm{g} / \mathrm{cm}^{3}(1 \mathrm{Cal}=4184 \mathrm{J})\)

Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of _____. $$v(t)=2 t+6, \text { for } 0 \leq t \leq 8$$
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