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Chapter 7: Problem 120

Find the exact area. Under \(f(x)=1 /(x+1)\) between \(x=0\) and \(x=2\)

Short Answer

Expert verified
The exact area under the curve is \( \ln 3 \).

Step by step solution

01

Understand the Problem

We need to find the area under the curve of the function \( f(x) = \frac{1}{x+1} \) from \( x=0 \) to \( x=2 \). This requires integrating the function over the given interval.
02

Set Up the Integral

To find the area under a curve represented by a function \( f(x) \) from \( x=a \) to \( x=b \), we compute the integral: \[A = \int_{a}^{b} f(x) \, dx\]For this problem, the integral is: \[A = \int_{0}^{2} \frac{1}{x+1} \, dx\]
03

Integrate the Function

The integral of \( \frac{1}{x+1} \) is given by the natural logarithm function: \[\int \frac{1}{x+1} \, dx = \ln|x+1| + C\]
04

Evaluate the Definite Integral

We now evaluate the integral from \( x = 0 \) to \( x = 2 \):\[A = \left[ \ln|x+1| \right]_{0}^{2} = \ln|2+1| - \ln|0+1|\]Simplifying, this becomes:\[A = \ln 3 - \ln 1\]
05

Simplify the Expression

Since \( \ln 1 = 0 \), the expression simplifies to:\[A = \ln 3\]

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Key Concepts

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

Definite Integral
A definite integral is a fundamental concept in calculus, used to calculate the accumulated value of a function over a specific interval. Unlike an indefinite integral, which accounts for a family of antiderivatives, a definite integral provides a numerical value that represents the area under a curve. This area is bounded by the curve itself, the x-axis, and the vertical lines at the interval's boundaries.

When evaluating definite integrals, we follow these main steps:
  • Identify the function to integrate, as well as the limits of integration.
  • Find the antiderivative of the function, and then apply the Fundamental Theorem of Calculus.
  • Subtract the value of the antiderivative at the lower limit from its value at the upper limit.
This technique is especially useful for finding physical quantities such as areas, volumes, and accumulated values. Getting comfortable with definite integrals helps in understanding complex mathematical problems.
Area Under a Curve
Finding the area under a curve is a classic application of integral calculus. It involves using definite integrals to calculate the total area captured by a curve between given bounds on the x-axis.

The curve we consider is expressed as a function, and the area is simply the space that lies between this curve and the x-axis within a certain interval. Here鈥檚 how it works:
  • First, we define the function that describes the curve.
  • Then, we specify the interval on the x-axis.
  • We set up the definite integral with these parameters.
  • Finally, we evaluate the integral, which results in the area under the curve.
Calculating this area helps in various fields such as physics, engineering, and economics, where it's essential to measure total quantities over periods or regions.
Natural Logarithm Function
The natural logarithm function, denoted as \(\ln(x)\), is the inverse of the exponential function \(e^x\). It has properties that make it especially useful in integral calculus, often surfacing when integrating functions of the form \(\frac{1}{x}\) or \(\frac{1}{x + c}\).

Here are some features of the natural logarithm:
  • It is defined only for positive real numbers.
  • Its base, \(e\), is approximately 2.71828 and is known as Euler's number.
  • It simplifies calculations involving growth processes and decays.
  • In integrals, \(\ln|x+1|\) arises as an antiderivative of \(\frac{1}{x+1}\), as seen in the original exercise problem.
Understanding the natural logarithm is crucial for integrating functions, solving differential equations, and analyzing variations in mathematical modeling.

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Most popular questions from this chapter

Decide whether the statements are true or false. Give an explanation for your answer.. The trapezoid rule approximation is never exact.

Find a substitution \(w\) and constants \(k, n\) so that the integral has the form \(\int k u^{n} d w\). $$\int \frac{\cos t}{\sin t} d t$$$$\int \frac{\cos t}{\sin t} d t$$

Find the exact area of the regions.Bounded by \(y=x^{2} / \sqrt{1-x^{2}}, y=0, x=0, x=1 / 2\).

Explain what is wrong with the statement. If \(a>0,\) then \(\int 1 /\left(x^{2}+4 x+a\right) d x\) always involves arctan.

Find the exact area of the regions.Bounded by \(y=\left(3 x^{2}+x\right) /\left(\left(x^{2}+1\right)(x+1)\right)\) \(y=0, x=0, x=1\).
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