Problem 10 - Thomas Calculus

Question

Suppose that \(f\) and \(h\) are integrable and that $$ \int_{1}^{9} f(x) d x=-1, \int_{7}^{9} f(x) d x=5, \quad \int_{7}^{9} h(x) d x=4 $$ Use the rules in Table 5.6 to find $$ \begin{array}{ll}{\text { a. } \int_{1}^{9}-2 f(x) d x} & {\text { b. } \int_{7}^{9}[f(x)+h(x)] d x} \\ {\text { c. } \int_{7}^{9}[2 f(x)-3 h(x)] d x} & {\text { d. } \int_{9}^{1} f(x) d x} \\ {\text { e. } \int_{1}^{7} f(x) d x} & {\text { f. } \int_{9}^{7}[h(x)-f(x)] d x}\end{array} $$

Step-by-Step Solution

Answer

a. 2, b. 9, c. -2, d. 1, e. -6, f. 1

Step 1: Calculate Integral a

The problem is to find \( \int_{1}^{9} -2f(x) dx \). We can apply the constant multiple rule from Table 5.6, which states that \( \int_{a}^{b} c \cdot f(x) \, dx = c \cdot \int_{a}^{b} f(x) \, dx \). Thus, \( \int_{1}^{9} -2f(x) dx = -2 \cdot \int_{1}^{9} f(x) \ dx = -2 \cdot (-1) = 2 \).

Step 2: Calculate Integral b

The problem is to find \( \int_{7}^{9} [f(x) + h(x)] \, dx \). We use the sum rule for integrals from Table 5.6, which states \( \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \). Therefore, \( \int_{7}^{9} [f(x) + h(x)] dx = \int_{7}^{9} f(x) \, dx + \int_{7}^{9} h(x) \, dx = 5 + 4 = 9 \).

Step 3: Calculate Integral c

The problem is to find \( \int_{7}^{9} [2f(x) - 3h(x)] \, dx \). We again use the sum and constant multiple rules, which gives us: \( \int_{7}^{9} 2f(x) \, dx - \int_{7}^{9} 3h(x) \, dx = 2 \int_{7}^{9} f(x) \, dx - 3 \int_{7}^{9} h(x) \, dx \). Substituting the given values: \( 2 \times 5 - 3 \times 4 = 10 - 12 = -2 \).

Step 4: Calculate Integral d

The problem asks for \( \int_{9}^{1} f(x) \, dx \). Using the property about reversing limits of integration, we know \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \). Therefore, \( \int_{9}^{1} f(x) \, dx = - \int_{1}^{9} f(x) \, dx \). From the given, \( \int_{1}^{9} f(x) \, dx = -1 \), thus, \( \int_{9}^{1} f(x) \, dx = -(-1) = 1 \).

Step 5: Calculate Integral e

To find \( \int_{1}^{7} f(x) \, dx \), note the property \( \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \) for any number \( c \) between \( a \) and \( b \). Thus, we have \( \int_{1}^{7} f(x) \, dx = \int_{1}^{9} f(x) \, dx - \int_{7}^{9} f(x) \, dx \). Substituting the known values, this becomes \( -1 - 5 = -6 \).

Step 6: Calculate Integral f

For \( \int_{9}^{7} [h(x) - f(x)] \, dx \) use the sum rule again: \( \int_{9}^{7} h(x) \, dx - \int_{9}^{7} f(x) \, dx \). By the limits reversal property \( \int_{9}^{7} g(x) \, dx = -\int_{7}^{9} g(x) \, dx \), so \( - \int_{7}^{9} h(x) \, dx + \int_{7}^{9} f(x) \, dx \). Substituting, \( -4 + 5 = 1 \).

Key Concepts

Integration Rules

In integral calculus, integration rules help us solve integrals with ease and efficiency. Some of the most important rules include the constant multiple rule, the sum/difference rule, and the reversal of limits. These rules simplify complex integrals by allowing you to break them into simpler parts.
For example, the constant multiple rule states that multiplying a function by a constant within an integral means you can also pull that constant outside the integral, thus simplifying the computation. This becomes useful when dealing with problems where functions within integrals are scaled by constants. Similarly, the sum and difference rules allow you to split integrals that contain sums or differences of functions into separate integrals, which can often be more manageable.

• This way, you can focus on each function individually.
• It significantly reduces the complexity of the solution.

Understanding these rules is paramount for anyone learning integral calculus as they lay the foundation for more advanced integration techniques.

Definite Integrals

Definite integrals are used to find the exact value of the area under a curve, between two specified limits. Unlike indefinite integrals, which represent a family of functions, definite integrals provide a specific numerical result. This is indicated by the limits of integration, namely the lower and upper limits that bound the area under consideration.
The concept can be visualized as slicing the region between the curve, the x-axis, and the two vertical boundaries defined by these limits–often resembling the area of a geographical region between specified longitudes.

• For instance, evaluating a definite integral is akin to finding the total accumulation of change over a particular interval.
• Applications abound, such as in physics where they help determine the total distance traveled given a velocity-time graph.

Mastering definite integrals is an essential skill when working with problems involving real-world quantities and their continuous changes over specified intervals.

Reversing Limits of Integration

An interesting property of integrals is the ability to reverse the limits of integration. If you reverse the upper and lower limits, the integral’s sign changes. This property is useful when dealing with problems where the natural limits may have been provided in reverse, or when breaking a problem into simpler segments.
The math behind this property is straightforward. For an integral \( \int_{a}^{b} f(x) \, dx \), by switching the limits, you have \( \int_{b}^{a} f(x) \, dx = -\int_{a}^{b} f(x) \, dx \).

• This means the direction of measuring the area is changed.
• It’s a helpful property to remember when manipulating integrals.

With this understanding, complex integration problems become easier to handle, especially when aligning limits aligns better with known values or simplifies calculations.

Constant Multiple Rule

The constant multiple rule is an incredibly handy aspect of integral calculus. It asserts that multiplying a function by a constant alters the integral's value in a straightforward way. Specifically, \( \int_{a}^{b} c \cdot f(x) \, dx = c \cdot \int_{a}^{b} f(x) \, dx \). This indicates that you can "pull out" the constant during integration, focusing on the function itself.
This rule holds immense value because it simplifies many tasks by reducing potential sources of error. Instead of complicating the integration, it emphasizes the function's core structure.

• For example, calculating \( \int_{1}^{9} -2f(x) \ dx \) merely becomes \(-2 \cdot (-1) \), simplifying to \(2\).
• It allows you to handle larger multiplicative constants without losing sight of the integral's essence.

Thus, leveraging this rule increases both efficiency and accuracy, proving beneficial for anyone delving into intricate integral evaluations.