Problem 77

Question

Find the error in the following "proof" that $0=1 .$ In $\int(1 / t) d t$, set $u=1 / t$ and $d v=d t .$ Then $d u=-t^{-2} d t$ and $u v=1$. Integration by parts gives $$ \int(1 / t) d t=1-\int(-1 / t) d t $$

Step-by-Step Solution

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Answer

The error is in the algebraic simplification of the integration by parts result.

1Step 1: Review the Integration by Parts Formula

Recall the integration by parts formula: \[ \int u \, dv = uv - \int v \, du. \] This helps break down complex integrals by choosing parts of the function to differentiate and integrate.

2Step 2: Identify and Check Assignments

According to the problem, we assign $ u = \frac{1}{t} $ and $ dv = dt $. Therefore, $ du = - \frac{1}{t^2} dt $ and $ v = t $. However, by substitution, we should have $ v $ as the integral of $ dt $, which is indeed $ t $, but care must be taken with the calculations.

3Step 3: Apply Integration by Parts

Substitute the assignments into the integration by parts formula: \[ \int \frac{1}{t} \, dt = \left( \frac{1}{t} \cdot t \right) - \int t \left( -\frac{1}{t^2} \right) dt. \] After processing the integrals, we get an integral setup which is simplified as \[ \int \frac{1}{t} \, dt = 1 + \int \frac{1}{t} \, dt. \] Notice a sign correction in the integral part.

4Step 4: Analyze the Resulting Integral

Bringing all terms to one side yields \[ 0 = 1. \] This is mathematically impossible and results from incorrect simplifications in earlier steps. Reevaluate the integration by parts to check for any algebraic or arithmetic errors, particularly focusing on handling negatives properly in subtraction.

Key Concepts

CalculusIntegration TechniquesAlgebraic Errors

Calculus

Calculus is a branch of mathematics that studies how things change. It is mainly divided into two parts: differentiation and integration. Differentiation focuses on finding the rate of change or the slope of a curve, whereas integration helps to find areas under curves or the accumulation of quantities. In this context, we are exploring integration, specifically through the technique known as Integration by Parts. This technique is derived from the product rule of differentiation and is a powerful tool to evaluate integrals that are difficult to tackle directly. Whenever you encounter a problem involving integration, especially with complex functions, calculus provides you with various tools to simplify and solve these problems. The interaction between terms as seen in the integration by parts is a prime example of applying calculus to strip down complex expressions into more manageable pieces.

Integration Techniques

Integration by Parts is one of several techniques in calculus used for integrating products of functions. The fundamental formula is: \[ \int u \, dv = uv - \int v \, du. \]To apply this technique effectively:

Choose the function u to be differentiated and the function dv to be integrated judiciously. A good rule of thumb is to let $ u $ be a function that becomes simpler upon differentiation and $ dv $ be a function whose integral is simple to find.
Determine $ du $ by differentiating $ u $.
Find $ v $ by integrating $ dv $.

In our original exercise, choices for $ u = \frac{1}{t} $ and $ dv = dt $ were made. Following these,

$ du = -\frac{1}{t^2} dt $,
$ v = t $,

should be handled with care to avoid errors, especially when negative signs are involved. Common errors occur when handling these derivatives and integrations, leading to incorrect results. Double-check calculations to catch sign changes and other missteps.

Algebraic Errors

Algebraic errors often slip into calculus problems, especially with integration techniques like Integration by Parts. Negative signs and incorrect simplifications are frequent pitfalls, as in the exercise where setting $ 0 = 1 $ was the result of an error.To avoid such errors:

Carefully follow through all steps of the integration and differentiation, particularly focusing on signs when subtracting expressions.
After computing each integration step, ensure that the results align logically.
Utilize parentheses as needed to correctly interpret the order of operations.

In the given problem, being attentive to $ du = -\frac{1}{t^2} dt $ and correctly applying it in the integration by parts formula is crucial. An overlooked negative sign can lead to compounded errors and misleading results, exemplified by the incorrect final statement $ 0 = 1 $.

Problem 73

Problem 78

Other exercises in this chapter

Problem 72

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Problem 73

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Problem 78

. Suppose that you want to evaluate the integral $$ \int e^{5 x}(4 \cos 7 x+6 \sin 7 x) d x $$ and you know from experience that the result will be of the form

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Problem 81

Show that $$ f(t)=f(a)+\sum_{i=1}^{n} \frac{f^{(i)}(a)}{i !}(t-a)^{i}+\int_{a}^{t} \frac{(t-x)^{n}}{n !} f^{(n+1)}(x) d x $$ provided that $f$ can be differen

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