Question

Determine whether or not each of the following functions is integrable over the proposed interval. Give a reason for your answer. (a) \(\cos \frac{1}{x},[0,1]\) (b) \(\frac{x^{2}+x-2}{x-1},[0,1]\) (c) \(\frac{x^{2}+x-2}{x-1},[0,2]\) (d) \(\frac{x^{2}+x+2}{x-1},[0,2]\)

Short answer

Question: Determine if the following functions are integrable over the given intervals: (a) \(\cos \frac{1}{x},[0,1]\) (b) \(\frac{x^{2}+x-2}{x-1}, [0,1]\) (c) \(\frac{x^{2}+x-2}{x-1}, [0,2]\) (d) \(\frac{x^{2}+x+2}{x-1}, [0,2]\) Answer: (a) Not integrable, due to an infinite discontinuity at x=0. (b) Integrable, equivalent to the continuous function \(x+2\) over \([0,1]\). (c) Integrable, equivalent to the continuous function \(x+2\) over \([0,2]\). (d) Not integrable, due to an infinite discontinuity at x=1.

Step-by-step solution

Check Continuity

Since \(\cos(\frac{1}{x})\) is continuous everywhere except x=0 (where it is undefined), we should only focus on x = 0.

Determine Integrability at x=0

Unfortunately, \(\cos(\frac{1}{x})\) oscillates infinitely as x approaches 0, which means it has an infinite discontinuity at x=0.

Conclusion

The function \(\cos(\frac{1}{x})\) is not integrable over the interval \([0,1]\) since it has an infinite discontinuity at x=0. (b) \(\frac{x^{2}+x-2}{x-1}, [0,1]\)

Factor the Expression

To analyze the continuity, let's factor the expression: \(\frac{(x+2)(x-1)}{(x-1)}\)

Simplify the Expression

Cancel the common terms in the numerator and denominator: \(x+2\)

Check Continuity

The simplified function \(x+2\) is continuous over the interval \([0,1]\).

Conclusion

The function \(\frac{x^{2}+x-2}{x-1}\) is integrable over the interval \([0,1]\) since it is equivalent to the continuous function \(x+2\). (c) \(\frac{x^{2}+x-2}{x-1}, [0,2]\)

Reuse Factored and Simplified Expression

From part (b), we know the simplified expression is \(x+2\).

Check Continuity

The simplified function \(x+2\) is continuous over the interval \([0,2]\).

Conclusion

The function \(\frac{x^{2}+x-2}{x-1}\) is integrable over the interval \([0,2]\) since it is equivalent to the continuous function \(x+2\). (d) \(\frac{x^{2}+x+2}{x-1}, [0,2]\)

Find the Discontinuity

This function has a discontinuity at x=1 (where the denominator is zero).

Determine Integrability at x=1

Since the polynomial in the numerator has no common factors with the denominator, the given function has an infinite discontinuity at x=1.

Conclusion

The function \(\frac{x^{2}+x+2}{x-1}\) is not integrable over the interval \([0,2]\) since it has an infinite discontinuity at x=1.

Key concepts

Infinite Discontinuity

When examining a function like \(\cos \frac{1}{x}\), on the interval \[0,1\], an important aspect to consider is whether the function has any discontinuities. An infinite discontinuity occurs when the function's values approach infinity as the input approaches a certain point. In our case, as x gets closer to 0, the function oscillates more rapidly, and the values of the cosine function do not settle into a finite number. This rapid oscillation means the function does not have a finite limit at that point.

As a result, the presence of an infinite discontinuity in a function over an interval is a critical barrier to integrability. In the realm of integral calculus, a function must be bounded and not have any infinite discontinuities within the interval of integration to be considered integrable. This is because the integral essentially represents a sum of infinitely many infinitely small quantities, and if any of those quantities are infinite, the sum cannot be finite.

Function Continuity

Continuity is foundational to understanding why certain functions like \(x^2 + x - 2)/(x-1)\) can be integrated over a given interval, such as \[0,1\] or \[0,2\]. A function is considered continuous at a point if the limit as you approach that point exists and is equal to the function's value at that point. Additionally, presence of continuity over an entire interval suggests there are no breaks, jumps, or asymptotes in that interval.

When a function can be simplified to eliminate a discontinuity—like canceling out corresponding terms in the numerator and denominator—this gap in the graph disappears, lending to a continuous, and thus integrable, function over the interval. For example, after simplifying \(x^2 + x - 2)/(x-1)\) to \(x + 2\), we noted this resulting function is continuous, paving the way for its integration within the specified ranges.

Integral Calculus

Integral calculus is the study of integration, a process that determines the total accumulation of a quantity and solves the problem of finding the area under the curve of a graph of a function. Integrals, in this regard, require functions to be integrable, which often means the functions must be continuous on the interval of integration. However, even when functions possess discontinuities, as long as they are not infinite, certain types of integrals—namely improper integrals—can still be used to find a value.

In our textbook examples, we find situations with both integrable and non-integrable functions. Some, like \(x^2 + x - 2)/(x-1)\) over \[0,1\] or \[0,2\], can be simplified and integrated, while others with infinite discontinuities, like \(\cos \frac{1}{x}\) over \[0,1\] or \(x^2 + x + 2)/(x-1)\) over \[0,2\], cannot be integrated over those intervals as the infinite values at points of discontinuity disrupt the requisite boundedness of a function's graph for a standard Riemann integral.