Problem 41 - Calculus Early Transcendentals

Question

Use only the fact that \(\int_{0}^{4} 3 x(4-x) d x=32\) and the definitions and properties of integrals to evaluate the following integrals, if possible. a. \(\int_{4}^{0} 3 x(4-x) d x\) b. \(\int_{0}^{4} x(x-4) d x\) c. \(\int_{4}^{0} 6 x(4-x) d x \) d. \(\int_{0}^{8} 3 x(4-x) d x\)

Step-by-Step Solution

Answer

Based on the step-by-step solution above, evaluate the following integrals: a. ∫(4 to 0) 3x(4-x) dx: Answer: -32 b. ∫(0 to 4) x(x-4) dx: Answer: -32 c. ∫(4 to 0) 6x(4-x) dx: Answer: 64 d. ∫(0 to 8) 3x(4-x) dx: Answer: Not possible

Step 1: Switch Limits of Integration

The property of integrals allows us to switch the limits of integration and negate the result of the original integral. So, we have: \(\int_{4}^{0} 3x(4-x) dx = -\int_{0}^{4} 3x(4-x) dx = -32\) b. \(\int_{0}^{4} x(x-4) dx\)

Step 2: Negate the Function

We can rewrite the integrand as \(x(4-x)(-1)\). Since it is the negative of the given integral's integrand, we just negate the result from the original integral. Therefore, \(\int_{0}^{4} x(x-4) dx = -\int_{0}^{4} 3x(4-x) dx = -32\) c. \(\int_{4}^{0} 6x(4-x) dx\)

Step 3: Combine Properties

First, rewrite the integrand as \(3x(4-x)(-2)\). Now, switch the limits of integration and negate the given integral's result, and then multiply by the 2. Thus, \(\int_{4}^{0} 6x(4-x) dx = 2\int_{0}^{4} 3x(4-x) dx = 2 \cdot 32 = 64\) d. \(\int_{0}^{8} 3x(4-x) dx\)

Step 4: Split the Integral

Since our original integral bounds are from 0 to 4, we can split this integral into two parts, with the first part's bounds being from 0 to 4 and the second part's bounds being from 4 to 8. \(\int_{0}^{8} 3x(4-x) dx = \int_{0}^{4} 3x(4-x) dx + \int_{4}^{8} 3x(4-x) dx\) The first part of the integral equals 32 because it is the original integral. However, we cannot evaluate the second part as it is not possible using the given information. So the final answer is not possible.

Key Concepts

Definite Integrals

A definite integral is a fundamental concept in calculus representing the area under the curve of a function between two points on the x-axis. Formally, if you have a function f(x) and you want to find the area between this function and the x-axis from x=a to x=b, you calculate the definite integral of f(x) with limits of integration from a to b, denoted as \( \int_{a}^{b} f(x) \, dx \).

In the exercise, the given integral \( \int_{0}^{4} 3x(4-x) \, dx = 32 \) represents such an area between x=0 and x=4. This number, 32, is key as it serves as a reference for evaluating other integrals through various properties of integration.

Limits of Integration

The limits of integration define the interval over which you are integrating a function. They are the 'a' and 'b' in \( \int_{a}^{b} f(x) \, dx \). Properties of definite integrals allow for flexibility with these limits. For instance, if you reverse the limits of integration, the value of the integral changes sign but not magnitude.

In step 1 of the solution for part (a), we switch the limits from \( \int_{4}^{0} 3x(4-x) \, dx \) to \( \int_{0}^{4} 3x(4-x) \, dx \), changing the sign of the result. This demonstrates how understanding the properties of limits can help simplify the evaluation of integrals.

Linearity of Integration

The linearity of integration property states that the integral of a sum of functions is the sum of the integrals of those functions, and similarly, a constant multiplied by a function can be pulled out of the integral. More formally, for constants a and b and functions f(x) and g(x):

• \( \int (af(x) + bg(x)) \, dx = a \int f(x) \, dx + b \int g(x) \, dx \)

We apply this property in parts (b) and (c). In (b), we factor out a -1 to find the negative of the given integral, while in (c), we factor out -2, switch the limits, and then apply the given integral's value, neatly showcasing the linearity property in action.

Integral Symmetry Properties

Integral symmetry properties involve understanding the behavior of functions and how they relate to their integrals. If a function is even, meaning \( f(x) = f(-x) \), its integral over a symmetric interval [−a, a] simplifies to twice the integral from [0, a]. If a function is odd, meaning \( f(x) = -f(-x) \), its integral over a symmetric interval [−a, a] is zero.

While symmetry properties are not directly applied in this exercise due to the lack of symmetry in the intervals and functions, understanding these properties is crucial for evaluating more complex integrals efficiently.