/*! 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 39 Suppose the graph of \(f\) on th... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose the graph of \(f\) on the interval \([a, b]\) has length \(L,\) where \(f^{\prime}\) is continuous on \([a, b] .\) Evaluate the following integrals in terms of \(L\) a. \(\int_{a / 2}^{b / 2} \sqrt{1+f^{\prime}(2 x)^{2}} d x \quad\) b. \(\int_{a / c}^{b / c} \sqrt{1+f^{\prime}(c x)^{2}} d x\) if \(c \neq 0\)

Short Answer

Expert verified
(a) \(\int_{a / 2}^{b / 2} \sqrt{1+f^{\prime}(2 x)^{2}} d x\) (b) \(\int_{a / c}^{b / c} \sqrt{1+f^{\prime}(c x)^{2}} d x\), where \(c \neq 0\) Answer: (a) \(\frac{1}{2}L\) (b) \(\frac{1}{c}L\)

Step by step solution

01

Definition of arc length

First, let's recall the formula for arc length, which is given as: $$L = \int_{a}^b \sqrt{1 + (f'(x))^2} dx$$ Now, we will proceed to evaluate the integrals in terms of \(L\).
02

Solve integral (a)

First, we will evaluate the integral: $$\int_{a / 2}^{b / 2} \sqrt{1+f^{\prime}(2 x)^{2}} d x$$ To solve this integral, we will use the change of variables technique by setting \(u=2x\). Thus, \(du = 2dx\), and the integral becomes: $$\frac{1}{2}\int_{a}^{b} \sqrt{1+(f^{\prime}(u))^2} du$$ Comparing this integral to the arc length formula, we can see that this integral is just equal to \(\frac{1}{2}L\). So, the answer to integral (a) is: $$\frac{1}{2}L$$
03

Solve integral (b)

Next, we will evaluate the integral: $$\int_{a / c}^{b / c} \sqrt{1+f^{\prime}(c x)^{2}} d x$$ Again, we will use the change of variables technique. This time, we will set \(u=cx\). Thus, \(du = c dx\), and the integral becomes: $$\frac{1}{c}\int_{a}^{b} \sqrt{1+(f^{\prime}(u))^2} du$$ Comparing this integral to the arc length formula, we can see that this integral is just equal to \(\frac{1}{c}L\). So, the answer to integral (b) is: $$\frac{1}{c}L$$

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Key Concepts

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

Integral Calculus
Integral calculus is a branch of mathematics focussing on the accumulation of quantities and the overarching concept behind the calculation of areas and volumes. It’s a fundamental tool within the calculus domain along with its counterpart, differential calculus. The two main operations of integral calculus are the indefinite integral, which represents an antiderivative, and the definite integral, which gives the accumulation of a quantity.

For instance, if you’re looking to find how much area is under a curve between two points, or the total distance travelled by an object along a curve, you'll likely use the definite integral. In the context of arc length, we leverage definite integrals to quantify the length of the curve itself, which is intrinsic to understanding the given textbook exercise.
Arc Length Formula
The arc length formula is a quintessential application of integral calculus. It allows us to find the length of a smooth curve defined by a function f(x), over an interval [a, b].

Mathematically, the length L can be represented as:

\[ L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx \]
Where f'(x) is the derivative of the function f(x). This formula stems from the Pythagorean theorem and gives rise to the integral in the exercise at hand. When properly understood, the arc length formula empowers students to tackle a range of problems that involve measuring the extent of curves.
Derivative Continuity
Derivative continuity is a core concept in calculus, referring to the idea that a function’s derivative does not have any abrupt changes, holes, or jumps within a given interval. Continuity of the derivative is an essential criterion for many applications of calculus, including the ability to accurately measure arc length.

For the function to have a well-defined arc length over an interval [a, b], it is crucial that the derivative f'(x) is continuous on that interval. This ensures that the curve doesn't have any sharp turns or breaks that would complicate measurement. Hence, the condition in the exercise that the function's derivative is continuous is not arbitrary; it's a necessary condition to ensure that the calculation of arc length is possible.
Change of Variables Technique
The change of variables technique, also known as u-substitution, is an essential method in integral calculus for simplifying integrals. It is used when an integral contains a composite function and allows us to transform a difficult integral into a more manageable one.

For example, if the integral involves a function f'(cx), setting u = cx can help us rewrite the integral in terms of u, simplifying the calculation. This technique plays a pivotal role in solving the integrals in the textbook exercise, as it indicates that the change of variable not only simplifies the integral but also scales the arc length appropriately, relying on the arc length formula. The correct application of this technique directly leads to evaluating the given integrals with respect to the known length L, as demonstrated in the textbook solutions.

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Most popular questions from this chapter

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