Problem 24 - Applied Calculus

Question

Find the exact area. Between \(y=\ln x\) and \(y=\ln \left(x^{2}\right)\) for \(1 \leq x \leq 2\).

Step-by-Step Solution

Answer

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

Step 1: Understand the problem

We need to find the area between the curves defined by the equations \(y = \ln x\) and \(y = \ln(x^2)\) over the interval \([1, 2]\).

Step 2: Simplify \(y = \ln(x^2)\)

Simplifying \(y = \ln(x^2)\), we get \(y = 2\ln x\). Therefore, the area between the curves is between \(y = \ln x\) and \(y = 2\ln x\).

Step 3: Set Up the Integral

The area \(A\) between two curves can be found using the integral formula \(A = \int_{a}^{b} (f(x) - g(x)) \, dx\), where \(f(x)\) is the upper function and \(g(x)\) is the lower function. Here, \(f(x) = 2\ln x\) and \(g(x) = \ln x\).

Step 4: Evaluate the Integral

Calculate the integral from 1 to 2:\[\int_{1}^{2} (2\ln x - \ln x) \, dx = \int_{1}^{2} \ln x \, dx\].

Step 5: Integrate \(\ln x\)

The integral of \(\ln x\) is found using integration by parts. Let \(u = \ln x\) and \(dv = dx\), then \(du = \frac{1}{x}dx\) and \(v = x\). Applying integration by parts:\[\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C\].

Step 6: Solve the Definite Integral

Apply the limits to the integral:\[\left[x \ln x - x\right]_{1}^{2} = \left[2 \ln 2 - 2\right] - \left[1 \ln 1 - 1\right]\].Since \(\ln 1 = 0\), it reduces to:\[ = (2 \ln 2 - 2) - (0 - 1) = 2 \ln 2 - 2 + 1 = 2 \ln 2 - 1\].

Step 7: Final Answer

The exact area between the two curves from \(x = 1\) to \(x = 2\) is \(2 \ln 2 - 1\).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that studies the accumulation of quantities. Fundamentally, it integrates many small pieces to determine whole size, area, volume, etc. Unlike differential calculus, which looks at rates of change, integral calculus focuses on understanding total quantities.

To calculate the area between two curves, we use definite integrals, which provide us with a method to calculate finite areas in a two-dimensional plane. The formula involves subtracting one function from another and then integrating over the desired interval. This process adds up all the small strips of area between the curves, giving the overall area.

In our problem, we're finding the area between curves given by two logarithmic functions: \( y = \ln x \) and \( y = \ln(x^2) \). With simplification, this turns into finding the difference between \( y = \ln x \) and \( y = 2\ln x \) on the interval \([1, 2]\).

Integration by Parts

Integration by parts is a method used to integrate products of functions. It's analogous to the product rule in differentiation but used in reverse for integrals. The integration by parts formula is:\[\int u \, dv = uv - \int v \, du\]where \( u \) and \( dv \) are parts of the integrand that you select strategically.

In our example, we need to integrate \( \ln x \). We choose \( u = \ln x \) because its derivative is simpler, and \( dv = dx \), which is already a basic form of integration.

• Calculate \( du = \frac{1}{x}dx \)
• Integrate \( dv = dx \) to get \( v = x \)

Substitute into the formula to get:\[\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C\]Here, \(C\) is the constant of integration, which vanishes in definite integrals.

Definite Integral

A definite integral computes the net area under a curve between two points \( a \) and \( b \). It provides a precise value rather than a formula like in an indefinite integral. The result accounts for all that lies above the x-axis and subtracts any area below it.

In the context of our problem, the definite integral \[ \int_{1}^{2} (2 \ln x - \ln x) \, dx = \int_{1}^{2} \ln x \, dx\] represents the total area between the defined bounds. By evaluating this integral, particularly using the results from integration by parts, we can substitute the bounds to compute the exact area.

• The antiderivative of \( \ln x \) is \( x \ln x - x \).
• Apply the limits: \([1, 2]\).

The computation is as follows:\[ [x \ln x - x]_{1}^{2} = (2 \ln 2 - 2) - (0 - 1) = 2 \ln 2 - 1 \]Thus, the area between the curves is \(2 \ln 2 - 1\).