/*! 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 67 Function defined as an integral ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Function defined as an integral Find the are length of the function \(f(x)=\int_{e}^{x} \sqrt{\ln ^{2} t-1} d t\) on \(\left[e, e^{3}\right]\).

Short Answer

Expert verified
The arc length is \(L = 2e^{3} - e\).

Step by step solution

01

Differentiate \(f(x)\) with respect to \(x\)

Using the Fundamental Theorem of Calculus, when differentiating an integral of the form \(f(x) = \int_{a}^{x} g(t) dt\), we get: $$\frac{df}{dx} = g(x)$$ So in our case, the derivative of \(f(x) = \int_{e}^{x} \sqrt{\ln ^{2} t-1} dt\) with respect to \(x\) is: $$\frac{df}{dx} = \sqrt{\ln ^{2} x-1}$$
02

Apply the arc length formula

Using the arc length formula, \(L = \int_{a}^{b} \sqrt{1 + (\frac{df}{dx})^2} dx\), with \(a = e\) and \(b = e^{3}\): $$L = \int_{e}^{e^{3}} \sqrt{1 + \left(\sqrt{\ln ^{2} x-1}\right)^2} dx$$ Simplify the expression under the square root: $$L = \int_{e}^{e^{3}} \sqrt{1 + (\ln^{2}x-1)} dx$$ $$L = \int_{e}^{e^{3}} \sqrt{\ln^{2}x} dx$$
03

Compute the definite integral

Now we need to find the integral to calculate the arc length: $$L = \int_{e}^{e^{3}} \sqrt{\ln^{2}x}\: dx$$ We can write the integrand as $$L = \int_{e}^{e^{3}} (\ln x)\: dx$$ Use substitution method to solve the integral: let \(u = \ln x\), then \(x = e^u\) and \(dx = e^u du\): $$L = \int_{1}^{3} ue^{u} du$$ Next, apply integration by parts with \(v=e^u\) and \(dw = udu\): $$L = \int_{1}^{3} u \: e^u du = \left[u e^u\right]_{1}^{3} - \int_{1}^{3} e^u du = \left[(3e^{3} - e^{3}) - (e^{3}-e)\right]$$ Finally, calculate the arc length: $$L = 2e^{3} - e$$ Thus, the arc length of the function \(f(x)=\int_{e}^{x} \sqrt{\ln ^{2} t-1} d t\) on \(\left[e, e^{3}\right]\) is \(L = 2e^{3} - e\).

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.

Integral Calculus
Integral calculus is a fundamental branch of calculus focused on the concept of an integral, which provides a way to calculate the accumulation of quantities, such as areas under curves, total distances covered, or volumes of solids. It is often seen as the counterpart to differential calculus, which studies the rates at which quantities change.

The process of finding integrals is called integration. There are several techniques to perform this, such as substitution, integration by parts, and partial fractions. In the context of finding arc length, integration plays a pivotal role in summing up infinitesimally small segments along a curve to obtain the overall length. To understand arc length, one needs to comprehend how integration accounts for continuous accumulation along a curve and how various integration techniques simplify the process of finding a solution to an integral.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus bridges the two central concepts of calculus: differentiation and integration, revealing how they are inverse operations. It consists of two related parts. The first part provides a certain integral to be an antiderivative of the function being integrated. Specifically, if you have a continuous function, \( f(x) \) and you are integrating it over a range from \( a \) to \( x \) in the form \( \int_{a}^{x} f(t) dt \), the derivative of this integral with respect to \( x \), is \( f(x) \) itself. This is why in the step-by-step solution provided, after differentiating the integral function, we get the original function under the square root as the derivative, \( f'(x) = \sqrt{\ln ^{2} x-1} \).

The second part, sometimes called the Evaluation Theorem, tells us that if we want to evaluate the definite integral of a continuous function from \( a \) to \( b \), we can find any antiderivative of the function and simply subtract its values at \( b \) and \( a \) to get the result. This is especially useful for calculating areas under curves or, as in our case, the arc length of a function.
Integration by Parts
Integration by parts is a technique used to integrate the product of two functions. It is derived from the product rule for differentiation and is formally stated by the equation \( \int u dv = uv - \int v du \). In practice, it involves choosing which function in the integral to set as \( u \) (which you differentiate), and \( dv \) (which you integrate), to simplify the computation.

In the exercise provided, this method is necessary to tackle the integral of \( u e^u \), since it involves the product of \( u \) and \( e^u \). To apply integration by parts, the student sets \( u \) to \( \ln x \) and differentiates it, while \( dv \) is the remaining part of the function, which in turn is integrated. The aim is to transform the original integral into a simpler form that can be easily evaluated, eventually leading to the expression \( 2e^{3} - e \) for the arc length.

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

Fastest descent time The cycloid is the curve traced by a point on the rim of a rolling wheel. Imagine a wire shaped like an inverted cycloid (see figure). A bead sliding down this wire without friction has some remarkable properties. Among all wire shapes, the cycloid is the shape that produces the fastest descent time (see the Guided Project The amazing cycloid for more about the brachistochrone property). It can be shown that the descent time between any two points \(0 \leq a < b \leq \pi\) on the curve is $$ \text { descent time }=\int_{a}^{b} \sqrt{\frac{1-\cos t}{g(\cos a-\cos t)}} d t $$ where \(g\) is the acceleration due to gravity, \(t=0\) corresponds to the top of the wire, and \(t=\pi\) corresponds to the lowest point or the wire. a. Find the descent time on the interval \([a, b]\) by making the substitution \(u=\cos t\) b. Show that when \(b=\pi\), the descent time is the same for all values of \(a\); that is, the descent time to the bottom of the wire is the same for all starting points.

Evaluate the following integrals. $$\int \sin ^{2} x \cos ^{4} x d x$$

Maximum path length of a projectile (Adapted from Putnam Exam 1940) A projectile is launched from the ground with an initial speed \(V\) at an angle \(\theta\) from the horizontal. Assume the \(x\) -axis is the horizontal ground and \(y\) is the height above the ground. Neglecting air resistance and letting \(g\) be the acceleration due to gravity, it can be shown that the trajectory of the projectile is given by $$ \begin{array}{l} y=-\frac{1}{2} k x^{2}+y_{\max }, \quad \text { where } k=\frac{g}{(V \cos \theta)^{2}} \\ \text { and } y_{\max }=\frac{(V \sin \theta)^{2}}{2 g} \end{array} $$ a. Note that the high point of the trajectory occurs at \(\left(0, y_{\max }\right)\) If the projectile is on the ground at \((-a, 0)\) and \((a, 0)\) what is \(a ?\) b. Show that the length of the trajectory (arc length) is $$ 2 \int_{0}^{a} \sqrt{1+k^{2} x^{2}} d x $$ c. Evaluate the arc length integral and express your result in terms of \(V, g,\) and \(\theta\) d. For a fixed value of \(V\) and \(g,\) show that the launch angle \(\theta\) that maximizes the length of the trajectory satisfies \((\sin \theta) \ln (\sec \theta+\tan \theta)=1\) e. Use a graphing utility to approximate the optimal launch angle.

Practice with tabular integration Evaluate the following integrals using tabular integration (refer to Exercise 77 ). a. \(\int x^{4} e^{x} d x \quad\) b. \(\int 7 x e^{3 x} d x\) c. \(\int_{-1}^{0} 2 x^{2} \sqrt{x+1} d x\) d. \(\int\left(x^{3}-2 x\right) \sin 2 x \, d x\) e. \(\int \frac{2 x^{2}-3 x}{(x-1)^{3}} d x\) f. \(\int \frac{x^{2}+3 x+4}{\sqrt[3]{2 x+1}} d x\) g. Why doesn't tabular integration work well when applied to \(\int \frac{x}{\sqrt{1-x^{2}}} d x \, ?\) Evaluate this integral using a different method.

What are the two general ways in which an improper integral may occur?

See all solutions

Recommended explanations on Math 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.