Question

Which of the following integrals are improper, and why? \((a) \int_{0}^{x} e^{-r t} d t\) \((d) \int_{-\infty}^{0} e^{r t} d t\) (b) \(\int_{2}^{3} x^{4} d x\) \((e) \int_{1}^{5} \frac{d x}{x-2}\) (c) \(\int_{0}^{1} x^{-2 / 3} d x\) \((f) \int_{-3}^{4} 6 d x\)

Short answer

The improper integrals are (c), (d), and (e), due to infinite discontinuity or infinite interval.

Step-by-step solution

Understanding Improper Integrals

An integral is considered improper if either the interval of integration is infinite, or the function has an infinite discontinuity (like a vertical asymptote) within the interval. This requires special methods for evaluation compared to proper integrals.

Analyze Integral (a)

The integral \( \int_{0}^{x} e^{-r t} d t \) has no infinite limits of integration and the function \( e^{-r t} \) is continuous over any finite interval, provided \( r \) is constant. This is a proper integral.

Analyze Integral (b)

The integral \( \int_{2}^{3} x^{4} d x \) has finite limits from 2 to 3, and the function \( x^4 \) is continuous over this interval. Thus, it is a proper integral.

Analyze Integral (c)

The integral \( \int_{0}^{1} x^{-2 / 3} d x \) involves the function \( x^{-2/3} \), which has a vertical asymptote at \( x = 0 \). This makes the integral improper due to the infinite discontinuity at the lower limit of integration.

Analyze Integral (d)

The integral \( \int_{-\infty}^{0} e^{r t} d t \) has an infinite lower bound, making it an improper integral due to the limit of integration extending to negative infinity.

Analyze Integral (e)

The integral \( \int_{1}^{5} \frac{d x}{x-2} \) has a vertical asymptote at \( x = 2 \) within the interval of integration \([1, 5]\). This causes the integral to be improper as the function becomes undefined at \( x = 2 \).

Analyze Integral (f)

The integral \( \int_{-3}^{4} 6 d x \) is over a finite interval and the function 6 is continuous everywhere on this interval. Thus, it is a proper integral.

Key concepts

Infinite Discontinuity

An infinite discontinuity in a function occurs when the function approaches infinity at certain points within the region of integration. This usually happens at vertical asymptotes, where the function grows without bounds as it nears a specific value of the variable. In the context of integrals, this makes the integration process improper, requiring special handling unlike with regular calculus integrals. When faced with an infinite discontinuity, such as in the case of the integral \( \int_{0}^{1} x^{-2 / 3} d x \), the function \( x^{-2/3} \) becomes infinitely large as \( x \) approaches 0 from the right. This point at \( x = 0 \) forces the need for different techniques to evaluate the integral, such as limits, to manage the behavior of the function near this point. Thus, integrals featuring infinite discontinuities must be handled with care to ensure accuracy and to obtain meaningful results. In conclusion, when identifying improper integrals, always check for infinite discontinuities within the limits of integration to ascertain the need for special evaluation methods.

Integration Techniques

To solve improper integrals, especially those with infinite discontinuities or infinite bounds, special integration techniques are employed. One common approach is using limits to redefine the integral around the problem feature. This way, you can evaluate the behavior of an integral as it approaches an undefined form. For instance, when dealing with the integral \( \int_{1}^{5} \frac{d x}{x-2} \), there is a vertical asymptote at \( x = 2 \). Here, you might split the integral at \( x = 2 \), examining the limits as \( x \) approaches this point from both directions. Techniques like substitution can sometimes facilitate the process by transforming the integral into a more manageable form. Another essential technique is known as splitting the interval of integration if the challenge is localized. For improper integrals with infinite limits, replacing these limits with a parameter and analyzing the limit of convergence as this parameter grows provides a path to evaluate the integral properly. Mastering these integration techniques is crucial in handling improper integrals efficiently and achieving accurate evaluations, which are not possible using standard integration methods alone.

Infinite Limits of Integration

When dealing with integrals that stretch over infinite bounds or approach infinity, the limits of integration themselves can make the integral improper. This occurs when either the lower or upper limit, or both, extend to infinity or negative infinity. In such cases, one cannot compute the integral using traditional methods due to the unbounded nature of the interval.For example, consider the integral \( \int_{-\infty}^{0} e^{r t} d t \), where the lower bound is negative infinity. This integral does not approach a finite limit on its own and would require redefinition using limits. This involves expressing the integral as a limit: \[ \text{Limit as } a \to -\infty \int_{a}^{0} e^{r t} d t \]. By doing this, we're able to study how the integral behaves as \( a \) approaches negative infinity, consequently providing a means to evaluate it beyond traditional methods.Understanding and managing integrals with infinite limits is vital to solving them accurately. By properly tackling the infinity aspect, these integrals offer insights into areas like probability, physics, and engineering, where such limitless boundaries commonly appear.