Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(b\) so that \(\int_{0}^{b} \ln x d x=0\).

Step-by-Step Solution

Answer

The value of \(b\) is \(e\).

Step 1: Set Up the Integral

The problem requires us to find the value of \(b\) such that the integral \(\int_{0}^{b} \ln x \, dx = 0\). This implies that the area under the curve \(y = \ln x\) from \(x = 0\) to \(x = b\) equals zero.

Step 2: Solve the Indefinite Integral

The integral \(\int \ln x \, dx\) can be solved using integration by parts. Let \(u = \ln x\) and \(dv = dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = x\). Using the integration by parts formula \(\int u \, dv = uv - \int v \, du\), we get:\[\int \ln x \, dx = x \ln x - \int x \left(\frac{1}{x}\right) \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C.\]

Step 3: Evaluate the Definite Integral

Now, evaluate the definite integral \(\int_{0}^{b} \ln x \, dx\). Using the result from Step 2, we have:\[\int_{0}^{b} \ln x \, dx = \left[ x \ln x - x \right]_{0}^{b} = \left( b \ln b - b \right) - \lim_{a \to 0^+} (a \ln a - a).\]

Step 4: Evaluate the Limit as \(a \to 0^+\)

In the expression \(\lim_{a \to 0^+} (a \ln a - a)\), note that as \(a \to 0^+\), the term \(a\) approaches zero. For the term \(a \ln a\), it is known from calculus that \(\lim_{a \to 0^+} a \ln a = 0\). Hence,\[\lim_{a \to 0^+} (a \ln a - a) = 0 - 0 = 0.\]

Step 5: Set the Evaluated Integral to Zero and Solve for \(b\)

Substitute the limit back into the evaluated integral:\[b \ln b - b - 0 = 0.\]Simplify to get:\[b \ln b = b.\]Divide both sides by \(b\) (assuming \(b eq 0\)):\[\ln b = 1.\]Solving for \(b\), we find:\[b = e^1 = e.\]

Step 6: Conclusion

Thus, the value of \(b\) that satisfies the integral \(\int_{0}^{b} \ln x \, dx = 0\) is \(b = e\).

Key Concepts

Integration by Parts

Integration by parts is a useful technique in calculus for solving integrals where the standard methods do not apply or seem complicated. It is particularly effective when dealing with the product of a polynomial and a logarithmic or exponential function.
This technique is inspired by the product rule for differentiation. The formula for integration by parts is given by:

• \[ \int u \, dv = uv - \int v \, du \]

In this formula, \(u\) and \(dv\) are parts of the integrand that we choose. As a rule of thumb, it helps to select \(u\) as a function that simplifies upon differentiation, and \(dv\) as what is left over.
To demonstrate, in the integral \( \int \ln x \, dx \), choose \(u = \ln x\) (since it simplifies to \(\frac{1}{x}\) upon differentiation), and \(dv = dx\). This strategy successfully simplifies the integral and is crucial for solving many complex calculus problems.

Natural Logarithm

The natural logarithm, denoted as \(\ln x\), is an important function in calculus. It is the inverse of the exponential function \(e^x\).
It is commonly used in integration, differentiation, and series expansions.

• For positive numbers, \(\ln x\) grows slower than other polynomial functions as \(x\) increases.
• The domain of the natural logarithm function is strictly positive numbers, as \(\ln x\) is undefined for non-positive values.

In calculus problems, the function \(\ln x\) frequently appears, making its behavior and properties essential to understand. For instance, in the integral \(\int \ln x \, dx\), appreciating the slow growth of \(\ln x\) helps in comprehending why the function and its integral are manageable using techniques like integration by parts.

Calculus Limits

Limits in calculus provide a way to approximate values and solve expressions that can't be easily calculated directly. They also help in understanding behavior of functions at points that are not explicitly defined.
In this context, we evaluate the limit

• \[ \lim_{a \to 0^+} (a \ln a - a) \]

As \(a \to 0^+\), the term \(a\) approaches zero, simplifying the expression significantly. Calculus provides us with the powerful concept that \(\lim_{a \to 0^+} a \ln a = 0\). This is vital in solving definite integrals where limits are part of the boundary conditions.
Mastering limits is critical in calculus as it underpins many more complex problems, such as finding derivatives and integrals. It is a foundational stone for understanding how functions behave at specific points, especially near discontinuities or undefined points.

Area Under a Curve

In calculus, the definite integral of a function over a defined interval is understood as the net area between the curve and the x-axis.
This understanding allows us to calculate quantities like displacement, work, and probability. For the function \(\ln x\), computing the definite integral \(\int_0^b \ln x \, dx\) is like finding the area it encompasses between the starting point \(x=0\) and the end point \(x=b\).
When the problem asks for this integral to equal zero, it means that the positive and negative areas must cancel each other out perfectly. Visualizing this can be complex due to the nature of \(\ln x\) near zero, where it trends towards negative infinity.
Ultimately, finding the specific \(b\) where this balance occurs deepens our comprehension of how a function's shape impacts integral values. It demonstrates how calculus beautifully bridges algebraic and geometric interpretations.