Problem 117 - University Calculus: Early Transcendentals

Question

A basic property of definite integrals is their invariance under translation, as expressed by the equation $$\begin{aligned}&\int_{a}^{b} f(x) d x=\int_{a-c}^{b-c} f(x+c) d x\\\ &(1)\end{aligned}$$ The equation holds whenever \(f\) is integrable and defined for the necessary values of \(x .\) For example in the accompanying figure, show that $$\int_{-2}^{-1}(x+2)^{3} d x=\int_{0}^{1} x^{3} d x$$ because the areas of the shaded regions are congruent. (GRAPH CAN'T COPY). Use a substitution to verify Equation (1)

Step-by-Step Solution

Answer

The integrals are equal, confirming the property using substitution.

Step 1: Understand the Integral Properties

We need to show that \( \int_{-2}^{-1}(x+2)^{3} \,dx = \int_{0}^{1} x^{3} \,dx \) by using the invariance under translation property. We are to prove this by substitution as suggested.

Step 2: Define the Substitution

Choose the substitution \( u = x + 2 \). This substitution aims to shift the function by \( c = 2 \), making it resemble Equation (1). Thus, the transformation is \( x = u - 2 \).

Step 3: Change the Limits of Integration

With the substitution \( u = x + 2 \), the limits change as follows:- When \( x = -2 \), \( u = -2 + 2 = 0 \).- When \( x = -1 \), \( u = -1 + 2 = 1 \).So the new integral limits are from 0 to 1.

Step 4: Substitute into the Integral

Substitute \( x = u - 2 \) and update the differential:\[ dx = du \]Thus, the integral becomes \( \int_{0}^{1} (u)^3 \, du \).

Step 5: Evaluate the Integral

Calculate \( \int_{0}^{1} u^3 \, du \):\[ \int u^3 \, du = \frac{u^4}{4} + C \]Evaluate from 0 to 1:\[ \left[ \frac{u^4}{4} \right]_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} \].

Step 6: Confirm the Calculation

Calculate \( \int_{0}^{1} x^3 \, dx \) directly to compare:\[ \int x^3 \, dx = \frac{x^4}{4} \]Evaluate from 0 to 1:\[ \left[ \frac{x^4}{4} \right]_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} \].Both evaluations return \( \frac{1}{4} \), confirming the invariance under translation.

Key Concepts

Invariance Under Translation

All definite integrals have a fascinating property known as the invariance under translation. This property allows us to shift the entire function along the x-axis without changing the value of the integral. Mathematically, the invariance under translation is expressed as:\[\int_{a}^{b} f(x) \, dx = \int_{a-c}^{b-c} f(x+c) \, dx\]Here's what it means:- The integral of a function over a specific interval can be recalculated by translating the function horizontally across the x-axis.- The new limits of integration shift by the same constant value as the function.- Despite the shift, the integral's calculated value remains the same.In terms of practical applications, this property is useful in simplifying definite integrals.Imagine you have an integral with a complicated function. If shifting it helps you work with an easier form, you can confidently do so. Just make sure the translation covers all values required for calculations.In our example problem, shifting helped change an integral of \((x+2)^3\) from \([-2,-1]\) to a much simpler integral of \(x^3\) from 0 to 1.

Substitution Method

The substitution method is a powerful tool for evaluating integrals. It helps transform complex integrals into simpler ones by changing the variable of integration. In our exercise, we used substitution to demonstrate the invariance under translation.Here's a step-by-step look:- **Choose a Substitution:** In the exercise, we set \(u = x + 2\). This transforms our function and its limits to an easier form.- **Change Limits of Integration:** We update our limits based on the substitution. When \(x = -2\), \(u = 0\). When \(x = -1\), \(u = 1\).- **Substitute in the Integral:** We replace \(x+2\) with \(u\) in the integral and \(dx\) with \(du\).This gives us a new integral to evaluate that's more straightforward:\[\int_{0}^{1} u^3 \, du\]By following these steps, you ensure the integral takes a form easier to manage or calculate. The substitution method is particularly helpful when dealing with products, powers, or composite functions.

Integrable Functions

Understanding integrable functions is essential for working with definite integrals. A function is considered integrable over an interval if its integral exists and provides a finite value.Some features and requirements include:- **Continuity:** Most functions that are continuous on a closed interval, say \([a, b]\), are integrable. Discontinuity needs special handling.- **Types of Functions:** Polynomial, exponential, logarithmic, and trigonometric functions are commonly integrable.In our example, \((x+2)^3\) is easily integrable over any interval, such as \([-2, -1]\) or \([0, 1]\). The procedure involves calculating the area under the curve between these points, which the substitution method helps simplify.Always verify that:- The function meets necessary conditions across the entire interval.- Singularities or undefined points within said interval might pose a problem but are often addressable with limits or special integration techniques.Recognizing an integrable function helps apply the correct method of integration, ensuring accurate results and a better understanding of the behavior of complex functions.