Evaluate Definite Integral ∫[4f(x) + 2g(x) - H(x)] Dx From -65 To -23

Hey guys! Today, we're diving into the fascinating world of definite integrals and how we can use their properties, specifically linearity, to solve some interesting problems. We're given three definite integrals with functions f(x), g(x), and h(x), all evaluated over the same interval from -65 to -23. Our mission, should we choose to accept it (and we totally do!), is to find the value of a new definite integral that combines these functions. Let's break it down step by step, making sure we understand each concept along the way. Are you ready to become a definite integral master? Let's jump in!

Understanding the Problem

Before we jump into the solution, let's make sure we understand what the problem is asking. We're given the following information:

$$\begin{array}{l} \int_{-65}^{-23} f(x) d x=5 \\ \int_{-65}^{-23} g(x) d x=18 \\ \int_{-65}^{-23} h(x) d x=28 \end{array}$$

These equations tell us the definite integral (the signed area under the curve) of each function f(x), g(x), and h(x) from x = -65 to x = -23. Now, the real question is this: What does the following integral equal?

$$\int_{-65}^{-23}[4 f(x)+2 g(x)-h(x)] d x = ?$$

This looks a bit more complex, right? We've got a combination of our original functions, each multiplied by a constant, and then added or subtracted. This is where the magic of linearity comes into play. So, let's explore the properties of definite integrals and see how they can help us crack this problem.

Properties of Definite Integrals: Linearity is Key

The key to solving this problem lies in understanding the properties of definite integrals, particularly the property of linearity. Linearity, in the context of integrals, essentially means that the integral of a sum (or difference) of functions is the sum (or difference) of their individual integrals, and that we can pull constant multiples out of the integral. Let's break that down into two main properties:

1. Sum/Difference Rule: The integral of a sum or difference of functions is equal to the sum or difference of the integrals of those functions. Mathematically, this can be expressed as:

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

   In simple terms, if you're integrating a function that's made up of adding or subtracting other functions, you can split the integral into separate integrals for each of those functions. This is super helpful because it allows us to deal with each piece of a complex function individually.

2. Constant Multiple Rule: The integral of a constant times a function is equal to the constant times the integral of the function. Mathematically:

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

   This means that if you have a constant multiplied by a function inside an integral, you can simply pull the constant out in front of the integral. This simplifies the integral and makes it easier to solve. Think of it like factoring out a common factor – it makes the whole expression cleaner and easier to work with.

These two properties are the bread and butter of dealing with integrals of combined functions. They let us break down complex integrals into simpler ones that we can handle more easily. In our case, we'll use these properties to separate the integral of 4f(x) + 2g(x) - h(x) into integrals of f(x), g(x), and h(x) individually, which we already know the values for. Pretty neat, huh?

Applying Linearity to Our Problem

Okay, now that we've got a solid grasp of linearity, let's put it into action and solve our problem! We're trying to find the value of:

$$\int_{-65}^{-23}[4 f(x)+2 g(x)-h(x)] d x$$

The first thing we're going to do is use the Sum/Difference Rule to split this integral into three separate integrals:

$$\int_{-65}^{-23} 4f(x) dx + \int_{-65}^{-23} 2g(x) dx - \int_{-65}^{-23} h(x) dx$$

See how we've taken the integral of the sum and difference and turned it into the sum and difference of integrals? Now, let's use the Constant Multiple Rule to pull the constants out in front of their respective integrals:

$$4 \int_{-65}^{-23} f(x) dx + 2 \int_{-65}^{-23} g(x) dx - \int_{-65}^{-23} h(x) dx$$

Look at that! It's starting to look a lot like the information we were given at the beginning. We've successfully transformed our original integral into a combination of integrals that we already know the values for. This is the power of linearity in action.

Plugging in the Known Values

Now comes the fun part: plugging in the values we know! Remember, we were given:

$$\begin{array}{l} \int_{-65}^{-23} f(x) d x=5 \\ \int_{-65}^{-23} g(x) d x=18 \\ \int_{-65}^{-23} h(x) d x=28 \end{array}$$

So, let's substitute these values into our expression:

$$4 \int_{-65}^{-23} f(x) dx + 2 \int_{-65}^{-23} g(x) dx - \int_{-65}^{-23} h(x) dx = 4(5) + 2(18) - (28)$$

All that's left to do now is some simple arithmetic. We're almost there!

Crunching the Numbers

Let's finish this off by performing the multiplication and subtraction:

$$4(5) + 2(18) - (28) = 20 + 36 - 28$$

$$20 + 36 - 28 = 56 - 28$$

$$56 - 28 = 28$$

So, after all that, we've arrived at our final answer! The value of the integral is 28.

The Grand Finale: The Answer and a Quick Recap

Alright, drumroll please... The value of the integral $\int_{-65}^{-23}[4 f(x)+2 g(x)-h(x)] d x$ is 28! Woohoo! We did it!

Let's take a moment to recap what we've done. We started with a seemingly complex integral involving a combination of functions. By understanding and applying the properties of linearity, specifically the Sum/Difference Rule and the Constant Multiple Rule, we were able to break down the integral into simpler parts. We then plugged in the given values for the individual integrals and performed the arithmetic to arrive at our final answer.

This problem perfectly illustrates the power of linearity in dealing with definite integrals. It's a fundamental concept that will come in handy time and time again as you delve deeper into calculus. So, remember these properties, practice applying them, and you'll be solving integral problems like a pro in no time!

Practice Makes Perfect: Try These Problems!

Now that you've seen how we can use linearity to solve definite integral problems, it's your turn to put your knowledge to the test! Here are a couple of practice problems to help you hone your skills:

Problem 1:

Given:

$$\begin{array}{l} \int_{0}^{3} p(x) d x= -2 \\ \int_{0}^{3} q(x) d x= 7 \\ \int_{0}^{3} r(x) d x= -10 \end{array}$$

Evaluate:

$$\int_{0}^{3}[3 p(x)-q(x)+5 r(x)] d x$$

Problem 2:

Given:

$$\begin{array}{l} \int_{1}^{5} u(x) d x= 12 \\ \int_{1}^{5} v(x) d x= -4 \end{array}$$

Evaluate:

$$\int_{1}^{5}[-2 u(x)+6 v(x)] d x$$

Work through these problems using the same steps we followed in the example. Remember to break down the integrals using linearity, pull out the constants, and then plug in the known values. Don't be afraid to make mistakes – that's how we learn! The key is to practice, practice, practice.

If you get stuck, don't worry! Review the steps we took in the main problem, and think about how you can apply those same principles to these new scenarios. You've got this! And who knows, you might even start to enjoy these integral puzzles. Good luck, and happy integrating!

Further Exploration: Beyond Linearity

While linearity is a powerful tool for evaluating definite integrals, it's just one piece of the puzzle. As you continue your journey in calculus, you'll encounter many other properties and techniques that can help you tackle even more complex integrals. Here are a few areas you might want to explore further:

• The Additive Interval Property: This property states that if a, b, and c are real numbers, then:

  $$\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$$

  In essence, it allows you to break an integral over a larger interval into the sum of integrals over smaller intervals. This is incredibly useful when dealing with piecewise functions or when you need to evaluate an integral in stages.

• Integration by Substitution (u-substitution): This technique is the reverse of the chain rule for differentiation. It's used to simplify integrals by substituting a part of the integrand with a new variable, u. This can often transform a complicated integral into a more manageable one.

• Integration by Parts: This technique is the reverse of the product rule for differentiation. It's used to integrate products of functions, and it's particularly helpful when dealing with integrals involving functions like xsin(x) or xe^(x).

• Trigonometric Integrals: These are integrals that involve trigonometric functions. They often require the use of trigonometric identities and special techniques to solve.

• Partial Fraction Decomposition: This technique is used to integrate rational functions (fractions where the numerator and denominator are polynomials). It involves breaking down the rational function into simpler fractions that are easier to integrate.

Exploring these topics will give you a more comprehensive understanding of integration and equip you with a wider range of tools for solving problems. Calculus is a vast and fascinating field, and there's always something new to learn!