https://file.aiquickdraw.com/mathbot/file/1726534710875p9ar4q8h.jpg

The task is to rewrite the improper integrals using the corresponding limit definition, without fully evaluating them. Let's go through each integral one by one:

### a) \(\int_0^2 \frac{dt}{2t - 1}\)
This integral has a potential singularity at \( t = \frac{1}{2} \), where the denominator becomes zero. Thus, this is an improper integral at \( t = \frac{1}{2} \).

We can split the integral into two parts:
\[
\int_0^2 \frac{dt}{2t - 1} = \int_0^{1/2} \frac{dt}{2t - 1} + \int_{1/2}^2 \frac{dt}{2t - 1}
\]
Each of these integrals is improper and can be rewritten using limits:
\[
\int_0^2 \frac{dt}{2t - 1} = \lim_{\epsilon \to \frac{1}{2}^-} \int_0^\epsilon \frac{dt}{2t - 1} + \lim_{\delta \to \frac{1}{2}^+} \int_\delta^2 \frac{dt}{2t - 1}
\]

### b) \(\int_{-\infty}^{\infty} e^{-x^2} dx\)
This is an improper integral due to the infinite limits of integration. To rewrite it using limits, we handle the infinity as:
\[
\int_{-\infty}^{\infty} e^{-x^2} dx = \lim_{A \to \infty} \int_{-A}^A e^{-x^2} dx
\]

### c) \(\int_{-3}^4 \frac{1}{x^2} dx\)
This integral has a singularity at \(x = 0\), where \( \frac{1}{x^2} \) becomes infinite. Hence, we split the integral at \(x = 0\) and rewrite it using limits:
\[
\int_{-3}^4 \frac{1}{x^2} dx = \int_{-3}^0 \frac{1}{x^2} dx + \int_0^4 \frac{1}{x^2} dx
\]
Rewriting using limits:
\[
= \lim_{\epsilon \to 0^-} \int_{-3}^{\epsilon} \frac{1}{x^2} dx + \lim_{\delta \to 0^+} \int_{\delta}^4 \frac{1}{x^2} dx
\]

### d) \(\int_0^1 \ln(x) dx\)
The natural logarithm \( \ln(x) \) has a singularity at \(x = 0\) since \(\ln(x)\) tends to \(-\infty\) as \(x \to 0\). Hence, the improper part is at \(x = 0\):
\[
\int_0^1 \ln(x) dx = \lim_{\epsilon \to 0^+} \int_{\epsilon}^1 \ln(x) dx
\]

### e) \(\int_0^1 \frac{\ln(x)}{(x - 1)^2} dx\)
Here, the integrand has two singularities: one at \(x = 0\) (due to \(\ln(x)\)) and one at \(x = 1\) (due to the denominator \((x - 1)^2\)). We split the integral at \(x = 1\) and handle both improper parts using limits:
\[
\int_0^1 \frac{\ln(x)}{(x - 1)^2} dx = \lim_{\epsilon \to 0^+} \int_{\epsilon}^{1-\delta} \frac{\ln(x)}{(x - 1)^2} dx + \lim_{\delta \to 0^+} \int_{1-\delta}^1 \frac{\ln(x)}{(x - 1)^2} dx
\]

---

These rewritten integrals now express the improper behavior at the boundaries or singular points.

Would you like any specific explanation of these steps or further clarification on a particular integral?

### Related Questions:
1. How do we evaluate improper integrals involving infinity?
2. What are the common techniques for dealing with singularities in integrals?
3. How can limits be applied to improper integrals with discontinuities?
4. What is the significance of splitting integrals at singularities?
5. How does one determine whether an improper integral converges or diverges?

**Tip:** When dealing with improper integrals, always identify potential points of singularity or infinity first. This will guide you in rewriting them properly with limits.

Rewrite the following improper integrals using the corresponding limit definition. You do not have to fully evaluate these integrals.

This guide explains how to rewrite improper integrals using limit definitions without fully evaluating them. Covers integrals with infinite limits and singularities, with examples such as ∫_0^2 1/(2t-1) dt and ∫_{-∞}^{∞} e^{-x^2} dx.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation