Problem 67 - Calculus Early Transcendentals

Question

Consider two functions \(f\) and \(g\) on [1,6] such that \(\int_{1}^{6} f(x) d x=10, \int_{1}^{6} g(x) d x=5\) \(\int_{4}^{6} f(x) d x=5,\) and \(\int_{1}^{4} g(x) d x=2 .\) Evaluate the following integrals. a. \(\int_{1}^{4} 3 f(x) d x \quad\) b. \(\quad \int_{1}^{6}(f(x)-g(x)) d x\) c. \(\int_{1}^{4}(f(x)-g(x)) d x \quad\) d. \(\int_{4}^{6}(g(x)-f(x)) d x\) e. \(\int_{4}^{6} 8 g(x) d x \quad\) f. \(\quad \int_{4}^{1} 2 f(x) d x\)

Step-by-Step Solution

Answer

In summary, we have found the following values for the integrals: a. \(\int_{1}^{4} 3 f(x) d x = 15\) b. \(\int_{1}^{6}(f(x)-g(x)) d x = 5\) c. \(\int_{1}^{4}(f(x)-g(x)) d x = 3\) d. \(\int_{4}^{6}(g(x)-f(x)) d x = -2\) e. \(\int_{4}^{6} 8 g(x) d x = 24\) f. \(\int_{4}^{1} 2 f(x) d x = -10 \)

Step 1: Constant factor rule

We have to find \(\int_{1}^{4} 3 f(x) d x\). We can apply the constant factor rule to get: \(3 \int_{1}^{4} f(x) d x\).

Step 2: Splitting the interval

To find the integral \(\int_{1}^{4} f(x) d x\), we can split it using the interval \([1,6]\). We know: \(\int_{1}^{6} f(x) d x = 10\) and \(\int_{4}^{6} f(x) d x = 5\) So, \(\int_{1}^{4} f(x) d x = \int_{1}^{6}f(x)d x - \int_{4}^{6}f(x)d x = 10 - 5 = 5\) Now, substituting back into the constant factor rule, the solution is: \(\int_{1}^{4} 3 f(x) d x = 3 \times 5 = 15\) #b. \(\int_{1}^{6}(f(x)-g(x)) d x\)#

Step 3: Sum rule

Using the sum rule, we get \(\int_{1}^{6}(f(x)-g(x)) d x = \int_{1}^{6} f(x) dx - \int_{1}^{6} g(x) dx\) We have both the values, \(\int_{1}^{6} f(x) dx = 10\) \(\int_{1}^{6} g(x) dx = 5\) Substituting the values, the solution is: \(\int_{1}^{6}(f(x)-g(x)) d x = 10 - 5 = 5\) #c. \(\int_{1}^{4}(f(x)-g(x)) d x\)#

Step 4: Sum rule

Following the sum rule like before, \(\int_{1}^{4}(f(x)-g(x)) d x = \int_{1}^{4} f(x) dx - \int_{1}^{4} g(x) dx\) We already know the values, \(\int_{1}^{4} f(x) d x = 5\) and \(\int_{1}^{4} g(x) d x = 2\) Substituting the values, the solution is: \(\int_{1}^{4}(f(x)-g(x)) d x = 5 - 2 = 3\) #d. \(\int_{4}^{6}(g(x)-f(x)) d x\)#

Step 5: Sum rule

Following the sum rule like before, \(\int_{4}^{6}(g(x)-f(x)) d x = \int_{4}^{6} g(x) dx - \int_{4}^{6} f(x) dx\) We already know the values, \(\int_{4}^{6} g(x) d x = 3\) and \(\int_{4}^{6} f(x) d x = 5\) Substituting the values, the solution is: \(\int_{4}^{6}(g(x)-f(x)) d x = 3 - 5 = -2\) #e. \(\int_{4}^{6} 8 g(x) d x\)#

Step 6: Constant factor rule

Following the constant factor rule, we get \(\int_{4}^{6} 8 g(x) d x = 8 \int_{4}^{6} g(x) d x\) We already know the value, \(\int_{4}^{6} g(x) d x = 3\) Substituting the values, the solution is: \(\int_{4}^{6} 8 g(x) d x = 8 \times 3 = 24\) #f. \(\int_{4}^{1} 2 f(x) d x\)#

Step 7: Constant factor rule and Reversal of limits

Using the constant factor rule and reversal of limits together, \(\int_{4}^{1} 2 f(x) d x = 2 \int_{4}^{1} f(x) d x = -2 \int_{1}^{4} f(x) d x\) We already know the value, \(\int_{1}^{4} f(x) d x = 5\) Substituting the values, the solution is: \(\int_{4}^{1} 2 f(x) d x = -2 \times 5 = -10\)

Key Concepts

Constant Factor Rule

Integral calculus often involves finding the area under a graph of a function. When a function is multiplied by a constant, this can significantly simplify the process of integration. The Constant Factor Rule in integral calculus is a fundamental property that states when a function is multiplied by a constant, the integral of this combination is the constant times the integral of the function alone.

For example, if we need to find the integral of a function like \(3f(x)\), we can take the constant \(3\) out of the integral. Mathematically, we express this as \(\int 3f(x)dx = 3\int f(x)dx\). This powerful rule allows us to handle more complex integrals by breaking them down into simpler parts, thus making our calculations and understanding of integral calculus easier and more efficient. When applied correctly, the Constant Factor Rule saves time and simplifies the integration process, allowing students to focus on more challenging parts of the problem.

Sum Rule

Integral calculus is not only about single functions; we often deal with sums or differences of functions. Here, the Sum Rule comes into play, providing an essential technique to integrate the sums or differences of functions separately. The Sum Rule tells us that the integral of a sum (or difference) of functions is equal to the sum (or difference) of their integrals.

In terms of notation, if we have two functions, \(f(x)\) and \(g(x)\), then the integral of their sum or difference is written as \(\int (f(x) \pm g(x))dx = \int f(x)dx \pm \int g(x)dx\). This rule is incredibly useful when faced with complex expressions, and it can significantly reduce the work needed to resolve complicated integrals. By applying the Sum Rule, students can approach their problems in a step-by-step manner, simplifying calculations and ensuring that complex integrals are more manageable.

Reversal of Limits

Sometimes in integral calculus, the order of our limits of integration may be reversed, leading to a situation that might seem confusing at first. The Reversal of Limits is a concept that addresses this scenario, and it is important for students to understand how it affects the integral's value. When we reverse the limits of integration, the resulting integral is the negative of the original integral.

This can be formally stated as \(\int_a^b f(x) dx = -\int_b^a f(x) dx\). Essentially, if you flip the upper and lower limits of an integral, the sign of the integral also flips. This property can be particularly helpful when calculating definite integrals over different intervals or when fixing an integral that was set up with the wrong bounds. Understanding the significance of the limits' order and mastering the Reversal of Limits principle can prevent mistakes and ensure clarity in solving integral calculus problems.