Question

Find the area between \(\ln x\) and the \(x\) -axis from \(x=1\) to \(x=2\).

Short answer

The area is \( 2 \ln 2 - 1 \).

Step-by-step solution

Understanding the Problem

We are asked to find the area between the curve of \( \ln x \) and the \( x \)-axis over the interval \( [1, 2] \). To find this area, we need to calculate the definite integral of \( \ln x \) over the interval \( [1, 2] \).

Setting Up the Integral

The area under a curve from \( a \) to \( b \) can be found using the definite integral \( \int_{a}^{b} f(x) \, dx \). In this case, \( f(x) = \ln x \), \( a = 1 \), and \( b = 2 \). Thus, we need to compute \( \int_{1}^{2} \ln x \, dx \).

Integration by Parts Setup

The function \( \ln x \) requires integration by parts. Recall the formula for integration by parts: \( \int u \, dv = uv - \int v \, du \). We set \( u = \ln x \), hence \( du = \frac{1}{x} \, dx \). Let \( dv = dx \), then \( v = x \).

Applying Integration by Parts

Substitute into the integration by parts formula: \[ \int \ln x \, dx = x \ln x - \int x \left( \frac{1}{x} \right) \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C \]. Now evaluate this from \( x=1 \) to \( x=2 \).

Evaluating the Definite Integral

First, calculate at \( x = 2 \): \( 2 \ln 2 - 2 \). Then calculate at \( x = 1 \): \( 1 \ln 1 - 1 = -1 \). Find the difference: \[ \left( 2 \ln 2 - 2 \right) - \left( -1 \right) = 2 \ln 2 - 2 + 1 = 2 \ln 2 - 1 \].

Conclusion

The area between the curve \( \ln x \) and the \( x \)-axis from \( x=1 \) to \( x=2 \) is \( 2 \ln 2 - 1 \).

Key concepts

Area Under Curve

When we talk about the "area under a curve," we're referring to the space between the curve and the x-axis over a certain interval. This area is a fundamental concept in calculus. To find it mathematically, we use an operation called integration, which allows us to sum up infinitely small vertical slices under the curve to get the total area.

In our problem, we look at the curve of the natural logarithm function, \( \ln x \). The interval given is from \( x = 1 \) to \( x = 2 \). This means we're only interested in the area lying between these two points on the x-axis. The area under the curve is significant because it helps us understand things like accumulation, which can apply to various real-world scenarios, from computing distance covered by a car to figuring out the amount of resource consumption over time.

• The definite integral is the tool we use for finding this area.
• We denote the area under a curve of function \( f(x) \) from \( a \) to \( b \) as \( \int_{a}^{b} f(x) \, dx \).
• In our specific case, we find \( \int_{1}^{2} \ln x \, dx \).

Simple enough, isn't it? Let's move on to learn a technique useful for evaluating this particular integral!

Integration by Parts

"Integration by Parts" is a technique derived from the product rule of differentiation, and it's used when we need to integrate the product of two functions. This method helps simplify integrals that appear complex at first glance. The general rule for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

In the problem, we need to use this technique because we're dealing with the function \( \ln x \), which requires this approach to integrate. Let's break down the setup for integration by parts:

• Choose \( u = \ln x \), as it's simpler to differentiate than to integrate.
• The derivative \( du = \frac{1}{x} \, dx \).
• For \( dv \), pick it to be \( dx \), since this allows us to simply integrate it to get \( v = x \).

Substituting into the formula, we compute:\[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx \].
By solving this, you find that the integral simplifies significantly. This clever approach is a lifesaver in calculus for solving a range of complicated integrals!

Natural Logarithm

The natural logarithm, often abbreviated as \( \ln \), is a special logarithm with base \( e \) (approximately 2.718). It's special because \( e \) serves as the natural base of logarithms in calculus and many areas of mathematics. Understanding \( \ln x \) is crucial because it recurrently appears in calculus, especially in integration and differential equations.

Here are some key characteristics of \( \ln x \):

• It is undefined for \( x \leq 0 \).
• \( \ln 1 = 0 \) because \( e^0 = 1 \).
• It increases without bound as \( x \) increases.

So, in our specific problem, we're exploring how this function behaves over the interval from \( x = 1 \) to \( x = 2 \). The natural logarithm isn't just about finding values; it's also about understanding growth rates.When you calculate areas or solve real-world problems, knowing how to integrate \( \ln x \) gives you the ability to apply the concept of exponential growth and decay in various scientific and financial computations.