Hitung Integral Tentu: 2f(x) + 5 Dx

Hey guys, welcome back to our math corner! Today, we've got a cool problem involving definite integrals and their properties. We're diving into a question that might look a bit intimidating at first glance, but trust me, once we break it down, it'll be a piece of cake. We're given some values for definite integrals of a function f(x) and asked to find the value of another definite integral. This is all about understanding how to manipulate and combine integral properties, so let's get started and make sure we nail this concept!

Understanding the Given Information

Alright, so the first thing we need to do is carefully look at what information is given to us. We have two key pieces of information about the function f(x):

1. ∫₀⁵ f(x) dx = 10: This tells us that the definite integral of f(x) from x=0 to x=5 is equal to 10. In simpler terms, the area under the curve of f(x) between 0 and 5 is 10.
2. ∫₃⁵ f(x) dx = 4: This one tells us that the definite integral of f(x) from x=3 to x=5 is 4. So, the area under f(x) between 3 and 5 is 4.

Our mission, should we choose to accept it, is to determine the value of ∫₀³ (2f(x) + 5) dx. This involves a slightly more complex integrand (the function inside the integral), which is 2f(x) + 5, and a different integration interval, from 0 to 3. To solve this, we'll need to leverage some fundamental properties of definite integrals. It's like having building blocks, and we need to figure out which blocks to use and how to stack them to reach our final answer. Remember, in mathematics, especially calculus, understanding the given conditions is paramount. It's the foundation upon which we build our solution. Don't just glance at the numbers; really understand what they represent in the context of integration. The interval of integration and the value of the integral are crucial pieces of information that guide us toward the correct approach. Without this careful initial observation, we might end up going down the wrong path, which is something we definitely want to avoid when tackling these kinds of problems. So, take a deep breath, read carefully, and let the given information sink in.

Key Properties of Definite Integrals We'll Use

Before we jump into solving, let's quickly recap some essential properties of definite integrals that will be our best friends for this problem. Knowing these by heart will make solving problems like this a breeze.

• Additivity Property (Interval Splitting): This is super important! It states that for any a, b, and c, we have: ∫ₐᶜ f(x) dx = ∫ₐᵇ f(x) dx + ∫<0xE2><0x82><0x99>ᶜ f(x) dx. This property allows us to break down an integral over a large interval into integrals over smaller, adjacent intervals. It's exactly what we need when we see intervals like [0, 5], [3, 5], and we want to find something over [0, 3]. Think of it as connecting the dots on a number line. If you want to go from point A to point C, you can go from A to B and then from B to C. The total distance (or integral) is the sum of the distances of the two smaller legs of the journey.

• Constant Multiple Rule: This one is also a lifesaver. It says that for any constant 'k', ∫ₐᵇ k * f(x) dx = k * ∫ₐᵇ f(x) dx. Basically, you can pull constants out of the integral. This will be useful when we deal with the 2f(x) part of our integrand.

• Sum/Difference Rule: This property allows us to integrate a sum or difference of functions separately: ∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx. This means we can split our integrand (2f(x) + 5) into two separate integrals.

• Integral of a Constant: The integral of a constant 'c' over an interval [a, b] is simply ∫ₐᵇ c dx = c * (b - a). This is how we'll handle the '+ 5' part of our integrand. The integral of a constant is just the constant multiplied by the length of the interval.

• Integral from a to b and b to a: While not directly used here, it's good to remember ∫ₐᵇ f(x) dx = - ∫<0xE2><0x82><0x99>ᵃ f(x) dx. This reverses the sign if you swap the limits of integration.

Mastering these properties is key to unlocking the solution. They are the tools in our calculus toolbox. Each property addresses a specific aspect of integral manipulation, and by combining them strategically, we can simplify complex integral expressions and find their values. It's like solving a puzzle; you have different shaped pieces (the properties), and you need to fit them together in the right order to reveal the complete picture. So, keep these rules handy as we move forward, and don't hesitate to revisit them if you feel stuck. They are your guideposts on the path to a correct answer. The additivity property, in particular, is often the lynchpin in problems where the given intervals don't directly match the desired interval, forcing us to find a way to connect them through intermediate points. This problem is a perfect example of where that property shines.

Step-by-Step Solution

Now, let's put those properties into action! Our goal is to find ∫₀³ (2f(x) + 5) dx. We're going to break this down piece by piece.

Step 1: Split the Integrand using the Sum Rule.

First, we can use the Sum Rule to separate the 2f(x) part and the 5 part:

∫₀³ (2f(x) + 5) dx = ∫₀³ 2f(x) dx + ∫₀³ 5 dx

This makes our problem much more manageable, as we can now tackle each integral separately.

Step 2: Handle the ∫₀³ 2f(x) dx part.

Using the Constant Multiple Rule, we can pull the 2 out:

∫₀³ 2f(x) dx = 2 * ∫₀³ f(x) dx

Now, the crucial part is figuring out the value of ∫₀³ f(x) dx. We're given ∫₀⁵ f(x) dx = 10 and ∫₃⁵ f(x) dx = 4. We need to find the integral from 0 to 3. This is where the Additivity Property comes in handy! We know that:

∫₀⁵ f(x) dx = ∫₀³ f(x) dx + ∫₃⁵ f(x) dx

We can rearrange this to solve for ∫₀³ f(x) dx:

∫₀³ f(x) dx = ∫₀⁵ f(x) dx - ∫₃⁵ f(x) dx

Plugging in the given values:

∫₀³ f(x) dx = 10 - 4

∫₀³ f(x) dx = 6

Awesome! So, now we know that ∫₀³ f(x) dx is 6. Let's substitute this back into our expression from Step 2:

2 * ∫₀³ f(x) dx = 2 * 6 = 12

So, the first part of our original integral evaluates to 12.

Step 3: Handle the ∫₀³ 5 dx part.

This is where the Integral of a Constant rule comes in. The constant is 5, and the interval is from 0 to 3. The length of the interval is (3 - 0) = 3.

∫₀³ 5 dx = 5 * (3 - 0)

∫₀³ 5 dx = 5 * 3

∫₀³ 5 dx = 15

So, the second part of our original integral evaluates to 15.

Step 4: Combine the results.

Now we just need to add the results from Step 2 and Step 3, based on our split in Step 1:

∫₀³ (2f(x) + 5) dx = (2 * ∫₀³ f(x) dx) + (∫₀³ 5 dx)

∫₀³ (2f(x) + 5) dx = 12 + 15

∫₀³ (2f(x) + 5) dx = 27

And there you have it! The value of the integral is 27. See? By breaking it down using the properties we discussed, it becomes much more straightforward. It's all about strategic application of these rules. Remember, the key was realizing we could find ∫₀³ f(x) dx by using the given information and the additivity property, and then applying the other rules for constants and sums. Keep practicing these steps, and you'll be a definite integral pro in no time!

Final Answer and Takeaways

So, after all that work, we've arrived at our final answer: 27. The value of ∫₀³ (2f(x) + 5) dx is indeed 27. This problem beautifully illustrates the power and flexibility of definite integral properties. We didn't need to know the actual function f(x); all we needed were the values of its integrals over specific intervals and a solid understanding of how to manipulate them.

What are the key takeaways here, guys?

• Don't get intimidated by complex integrands or intervals. Break them down! The Sum Rule and Constant Multiple Rule are your best friends for simplifying the function part.
• The Additivity Property (∫ₐᶜ = ∫ₐᵇ + ∫<0xE2><0x82><0x99>ᶜ) is your secret weapon when your desired interval isn't directly given but can be formed by combining or subtracting given intervals. In this case, it allowed us to find ∫₀³ f(x) dx from ∫₀⁵ f(x) dx and ∫₃⁵ f(x) dx.
• Always remember the integral of a constant: ∫ₐᵇ c dx = c * (b - a). It's a simple but essential rule for handling constant terms.

By combining these properties, we were able to solve the problem efficiently. It’s a fantastic example of how abstract mathematical rules translate into practical problem-solving techniques. Remember that each step was logical and built upon the previous one, guided by established calculus theorems. This methodical approach is crucial for tackling more advanced problems in the future. So, next time you see an integral problem, take a deep breath, identify the given information, recall the properties you know, and start breaking it down. You've got this! Keep practicing, and you'll find that these concepts become more intuitive with every problem you solve. Happy integrating, everyone!