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Magazine Chapter 7: Problem 33
Evaluate the given improper integral. \(\int_{0}^{1} \ln x d x\)
Short Answer
Expert verified
The integral evaluates to -1.
Step by step solution
01
Recognizing the Improper Integral
In this problem, we have to evaluate the integral \(\int_{0}^{1} \ln x\,dx\). This is considered an improper integral because \(\ln x\) becomes unbounded as \(x\) approaches 0. Hence, we will need to evaluate it as a limit.
02
Set Up the Limit for the Improper Integral
To handle the discontinuity at \(x = 0\), we set up the integral as a limit: \[ \lim_{a \to 0^{+}} \int_{a}^{1} \ln x \, dx \] This reformulation allows us to avoid the discontinuity at \(x = 0\) by integrating from \(a\) to 1 and then taking the limit as \(a\) approaches 0 from the positive side.
03
Integral of \(\ln x\)
To find \(\int \ln x \ dx\), we use integration by parts, where we let \(u = \ln x\) and \(dv = dx\). Thus, \(du = \frac{1}{x} dx\) and \(v = x\). Using the integration by parts formula \(\int u\, dv = uv - \int v\, du\), we have: \[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int dx = x \ln x - x + C \] Therefore, the antiderivative is \(x \ln x - x + C\).
04
Evaluate the Definite Integral from \(a\) to 1
Substitute the bounds into the antiderivative to evaluate the definite integral: \[ \int_{a}^{1} \ln x \, dx = \left[ x \ln x - x \right]_{a}^{1} = (1 \cdot \ln 1 - 1) - (a \cdot \ln a - a) \]Simplify: \[ = (0 - 1) - (a \ln a - a) = -1 - a \ln a + a \]
05
Take the Limit as \(a\) Approaches 0
We need to take the limit as \(a \to 0^{+}\) of the expression \(-1 - a \ln a + a\):\[ \lim_{a \to 0^{+}} (-1 - a \ln a + a) \]The term \(a \ln a\) approaches 0 as \(a \to 0^{+}\), since \(\ln a\) approaches \(-\infty\) but \(a\) approaches 0 faster, making \(a \ln a\) approach 0. The \(a\) term also approaches 0, leaving us with:\[ \lim_{a \to 0^{+}} (-1 + 0) = -1 \]
06
Final Conclusion
Thus, the value of the improper integral \(\int_{0}^{1} \ln x \, dx\) is \(-1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a handy technique used to integrate products of functions. It's based on the product rule for differentiation and can be summarized by the formula:\[ \int u \, dv = uv - \int v \, du \]In our example, to integrate \(\ln x\), we select:
- \(u = \ln x\)
- \(dv = dx\)
- \(du = \frac{1}{x} dx\)
- \(v = x\)
Limits in Calculus
Limits are essential in calculus, especially when dealing with improper integrals. They help in evaluating an integral over an interval where the function is not well-behaved or becomes infinite.For the integral \(\int_{0}^{1} \ln x \, dx\), \(\ln x\) is undefined at \(x = 0\). To manage this, we define the integral using a limit:\[ \lim_{a \to 0^{+}} \int_{a}^{1} \ln x \, dx \]With this expression, we bypass the discontinuity, allowing the limit \(a \to 0^{+}\) to approach the boundary smoothly. Grasping limits is crucial for handling discontinuities and infinite behavior in integrals.
Natural Logarithm Function
The natural logarithm function, \(\ln x\), is a basic yet powerful function in calculus with several unique properties. It is the inverse of the exponential function \(e^x\) and is defined for \(x > 0\).Here are some of its essential characteristics:
- \(\ln 1 = 0\) because \(e^0 = 1\).
- \(\ln x\) grows slowly compared to polynomial functions.
- The derivative of \(\ln x\) is \(\frac{1}{x}\), which is particularly useful for integration by parts.
Discontinuity Handling
Dealing with discontinuities is a critical part of solving improper integrals. A discontinuity can occur when the function is undefined or infinite at certain points, like \(\ln x\) at \(x = 0\) in our example.To manage this, we avoid direct evaluation and instead use limits. By formulating the problem as:\[ \lim_{a \to 0^{+}} \int_{a}^{1} \ln x \, dx \]we steer clear of the problematic point. Here's the breakdown:
- The term \(- a \ln a + a\) is crucial. As \(a \to 0^{+}\), though \(\ln a\) heads to \(-\infty\), \(a\) approaches zero even faster, making \(a \ln a\) go to 0.
- This insight ensures that including this limit process, instead of direct evaluation, leads to a meaningful and finite evaluation of the integral.
