/*! 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 2 Give a geometrical interpretatio... [FREE SOLUTION] | 91Ó°ÊÓ

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Give a geometrical interpretation of the function \(\ln x=\int_{1}^{x} \frac{d t}{t}\)

Short Answer

Expert verified
Answer: The geometrical interpretation of the logarithmic function \(\ln x\) is that it represents the accumulated area under the curve \(y=\frac{1}{t}\) from \(t=1\) to \(t=x\). This provides a direct connection between the logarithmic function and the integral expression.

Step by step solution

01

Geometrical representation of the integral

Integrals can be interpreted as the area under a curve. In this case, the integral \(\int_{1}^{x} \frac{dt}{t}\) represents the area under the curve \(y=\frac{1}{t}\) from \(t=1\) to \(t=x\).
02

Visualization of areas in the diagram

To better visualize the areas, one can draw a diagram where the curve \(y=\frac{1}{t}\) is plotted and the integration interval is represented by shading the region under the curve, between \(t=1\) to \(t=x\).
03

Connection between areas and logarithmic function

The integral \(\int_{1}^{x} \frac{dt}{t}\) calculates the total area under the curve \(y=\frac{1}{t}\) between \(1\) and \(x\). This total area is equal to the value of the logarithmic function \(\ln x\). Therefore, the logarithmic function can be interpreted as the accumulated area under the curve \(y=\frac{1}{t}\) from \(t=1\) to \(t=x\).
04

Geometrical interpretation summary

In conclusion, the geometrical interpretation of the logarithmic function \(\ln x\) is that it represents the accumulated area under the curve \(y=\frac{1}{t}\) from \(t=1\) to \(t=x\). This provides a direct connection between the logarithmic function and the integral expression.

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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 fundamental part of calculus focused on accumulation of quantities and the areas under curves. It involves integrals, which are mathematical expressions representing the accumulation of a quantity. For example, in our exercise, the integral \( \int_{1}^{x} \frac{dt}{t} \) calculates the area under the curve \( y = \frac{1}{t} \) from \( t = 1 \) to \( t = x \). This operation allows us to understand how quantities build up over an interval.

Integrals are often visualized as the total area beneath a curve and above the \( t \)-axis. When calculated, they provide a numerical value representing this area. This concept is crucial not only for solving mathematical problems but also for tackling real-world situations like finding distances, areas, and even volumes. Integrals serve as a bridge between geometry and algebra, offering a graphical insight into algebraic symbols and equations.
Logarithmic Function
The logarithmic function \( \ln x \) is a special type of function that is important in both mathematics and everyday applications, such as evaluating exponential growth. In the given exercise, \( \ln x \) is defined through an integral form: \( \ln x = \int_{1}^{x} \frac{dt}{t} \). This means that the natural logarithm of \( x \) can be interpreted as the total area under the curve \( y = \frac{1}{t} \) from \( t = 1 \) to \( t = x \).

A logarithmic function essentially answers the question: "To what power must we raise the base, \( e \), to obtain \( x \)?" Here, the base \( e \) is an irrational number approximately equal to 2.718. Logarithms help simplify multiplicative processes into additive ones, which can be incredibly useful for solving complex equations or real-life scenarios involving exponential growth.
  • They simplify the calculations of large multiplicative processes.
  • Used in sciences to deal with quantities that vary exponentially.
Area Under a Curve
The area under a curve is a graphical concept that integral calculus quantifies numerically. It involves calculating how much space is occupied beneath a given curve and above a specified interval on the \( x \)-axis.

In our particular scenario, the area under the curve \( y = \frac{1}{t} \) from \( t = 1 \) to \( t = x \) is significant because it defines the value of the logarithmic function \( \ln x \). By finding this area, we can connect the visual shape of the curve to a precise mathematical value.

This geometrical interpretation offers a deeper understanding of how functions like \( \ln x \) are formed and used. It shows how complex mathematical expressions can tell us about the continuous accumulation of space or growth over an interval. Recognizing these areas helps not only in pure math but also in fields like physics, engineering, and any discipline involving continuous change.

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Most popular questions from this chapter

Use the general slicing method to find the volume of the following solids. The solid with a semicircular base of radius 5 whose cross sections perpendicular to the base and parallel to the diameter are squares

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Archimedes' principle says that the buoyant force exerted on an object that is (partially or totally) submerged in water is equal to the weight of the water displaced by the object (see figure). Let \(\rho_{w}=1 \mathrm{g} / \mathrm{cm}^{3}=1000 \mathrm{kg} / \mathrm{m}^{3}\) be the density of water and let \(\rho\) be the density of an object in water. Let \(f=\rho / \rho_{w}\). If \(01,\) then the object sinks. Consider a cubical box with sides 2 m long floating in water with one-half of its volume submerged \(\left(\rho=\rho_{w} / 2\right) .\) Find the force required to fully submerge the box (so its top surface is at the water level).

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