Problem 10 - University Calculus: Early Transcendentals

Question

Suppose that \(f\) and \(h\) are integrable and that $$\int_{1}^{9} f(x) d x=-1, \quad \int_{7}^{9} f(x) d x=5, \quad \int_{7}^{9} h(x) d x=4$$ a. \(\int_{1}^{9}-2 f(x) d x\) b. \(\int_{7}^{9}[f(x)+h(x)] d x\) c. \(\int_{7}^{9}[2 f(x)-3 h(x)] d x \quad\) d. \(\int_{9}^{1} f(x) d x\) e. \(\int_{1}^{7} f(x) d x \$ f. \)\int_{9}^{7}[h(x)-f(x)] d x$

Step-by-Step Solution

Answer

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

Step 1: Compute Part (a)

Given: \[ \int_{1}^{9} f(x) \, dx = -1 \] We need to find \( \int_{1}^{9} -2 f(x) \, dx \). By the property of integrals (scaling property), we have:\[ \int_{a}^{b} c \cdot f(x) \, dx = c \int_{a}^{b} f(x) \, dx \]Therefore:\[ \int_{1}^{9} -2 f(x) \, dx = -2 \int_{1}^{9} f(x) \, dx = -2 \times (-1) = 2 \]

Step 2: Compute Part (b)

Given: \[ \int_{7}^{9} f(x) \, dx = 5, \quad \int_{7}^{9} h(x) \, dx = 4 \]We need to find:\[ \int_{7}^{9} [f(x) + h(x)] \, dx \]By the linearity of integrals,\[ \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \]Thus,\[ \int_{7}^{9} [f(x) + h(x)] \, dx = \int_{7}^{9} f(x) \, dx + \int_{7}^{9} h(x) \, dx = 5 + 4 = 9 \]

Step 3: Compute Part (c)

We need to find:\[ \int_{7}^{9} [2f(x) - 3h(x)] \, dx \]Use the linearity and scaling properties:\[ \int_{a}^{b} [cf(x) + dg(x)] \, dx = c \int_{a}^{b} f(x) \, dx + d \int_{a}^{b} g(x) \, dx \]Here:\[ c = 2, \quad d = -3 \]Thus:\[ \int_{7}^{9} [2f(x) - 3h(x)] \, dx = 2 \int_{7}^{9} f(x) \, dx - 3 \int_{7}^{9} h(x) \, dx \]\[ = 2 \times 5 - 3 \times 4 = 10 - 12 = -2 \]

Step 4: Compute Part (d)

Given:\[ \int_{1}^{9} f(x) \, dx = -1 \]We need to find:\[ \int_{9}^{1} f(x) \, dx \]The property of definite integrals states that reversing the limits of integration changes the sign:\[ \int_{b}^{a} f(x) \, dx = - \int_{a}^{b} f(x) \, dx \]Thus:\[ \int_{9}^{1} f(x) \, dx = - \int_{1}^{9} f(x) \, dx = -(-1) = 1 \]

Step 5: Compute Part (e)

We need to find:\[ \int_{1}^{7} f(x) \, dx \]Use the property of splitting the integral:\[ \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \]Thus:\[ \int_{1}^{9} f(x) \, dx = \int_{1}^{7} f(x) \, dx + \int_{7}^{9} f(x) \, dx \]Plug in given values:\[ -1 = \int_{1}^{7} f(x) \, dx + 5 \]Solve for \( \int_{1}^{7} f(x) \, dx \):\[ \int_{1}^{7} f(x) \, dx = -1 - 5 = -6 \]

Step 6: Compute Part (f)

We need to find:\[ \int_{9}^{7} [h(x) - f(x)] \, dx \]Reverse the limits to simplify using property:\[ \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \]Thus:\[ \int_{9}^{7} [h(x) - f(x)] \, dx = -\int_{7}^{9} [h(x) - f(x)] \, dx \]Now find:\[ \int_{7}^{9} [h(x) - f(x)] \, dx = \int_{7}^{9} h(x) \, dx - \int_{7}^{9} f(x) \, dx \]\[ = 4 - 5 = -1 \]So:\[ \int_{9}^{7} [h(x) - f(x)] \, dx = - (-1) = 1 \]

Key Concepts

Integration Properties

Understanding the properties of integrals is crucial when working with definite integrals. One of the fundamental rules is the **Scaling Property**. This property states that if you multiply a function by a constant, you can factor the constant out in front of the integral:

\[ \int_{a}^{b} c \cdot f(x) \, dx = c \int_{a}^{b} f(x) \, dx \]

This property simplifies calculations drastically, making complex integrations far more manageable.

Another important property is changing the **limits of integration**. When you reverse the order of the limits, the sign of the integral changes:

\[ \int_{b}^{a} f(x) \, dx = - \int_{a}^{b} f(x) \, dx \]

Lastly, the **Additivity Property** allows you to split the integral over an interval into the sum of two integrals over sub-intervals:

\[ \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \]

Incorporating these properties lets you solve integral problems more easily by breaking them down into simpler parts.

Calculus

Calculus, as a whole, is the mathematical study of continuous change, and integral calculus is a significant part of it. When dealing with definite integrals, you are finding the signed area under a curve within a specific interval.

Definite integration has many applications, from calculating areas and volumes to solving problems in physics and engineering. The integration process often involves finding an antiderivative, a function whose derivative gives you back the original function. This makes integration the reverse of differentiation.

Working with functions like polynomial or trigonometric functions often involves using established formulas and rules, but a deep understanding of integration can lead to more intuitive and creative problem-solving.

• Integration formulizes how we add up an infinite number of small quantities.
• It provides analytical tools for calculating precise values where counting would take infinitely long.

Mastering integral calculus involves familiarizing yourself with various techniques and properties, as used in problem-solving scenarios.

Linearity of Integration

Linearity is a key concept in integration, allowing many integrals to be simplified. The **Linearity of Integration** involves two main properties:

1. **Sum Property:** If you have a sum of two functions under the integral sign, you can split the integral into the sum of each function's integrals:

\[ \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \]

This property simplifies integration by letting you consider each function individually.

2. **Constant Factor Property:** You can factor constants out of the integral, as previously discussed. Combining these properties allows you to solve complex integrals more practically.

These principles form the bedrock for more complex integration strategies, like integration by parts or partial fraction decomposition. The Linearity of Integration is powerful because it makes otherwise complicated integrals much more accessible.