Definite Integral: Evaluating ∫(6r+5)^2 From 1 To -1

by ADMIN 53 views

Hey guys! Today, we're diving into a super interesting calculus problem: evaluating the definite integral of the function (6r+5)^2 from 1 to -1. This might seem a bit daunting at first, but don't worry, we'll break it down step by step so it's easy to understand. We're going to cover everything from expanding the integrand to applying the power rule and finally, evaluating the definite integral at the given limits. So, buckle up and let's get started!

Understanding the Problem

Before we jump into the solution, let's make sure we understand what we're trying to do. We're given the definite integral ∫(6r+5)^2 dr with limits of integration from 1 to -1. In simpler terms, we need to find the area under the curve of the function (6r+5)^2 between the points r = 1 and r = -1 on the r-axis. Remember, definite integrals give us a numerical value, unlike indefinite integrals which give us a function plus a constant.

The integral symbol ∫ tells us we're dealing with integration, and the function inside the integral, (6r+5)^2, is called the integrand. The 'dr' indicates that we're integrating with respect to the variable 'r'. The numbers 1 and -1 are the limits of integration, with 1 being the upper limit and -1 being the lower limit. These limits tell us the interval over which we're finding the area.

To solve this, we'll first need to expand the integrand, then find its antiderivative, and finally evaluate the antiderivative at the limits of integration. We'll use some basic algebraic techniques and the power rule for integration to get there. So, let’s move on to the first step: expanding the integrand.

Step 1: Expanding the Integrand

The first thing we need to do is expand the integrand, which is (6r+5)^2. This means we need to multiply (6r+5) by itself. We can use the FOIL method (First, Outer, Inner, Last) or the binomial theorem to do this. Let's use the FOIL method:

(6r + 5)^2 = (6r + 5)(6r + 5)

Now, we multiply the terms:

  • First: 6r * 6r = 36r^2
  • Outer: 6r * 5 = 30r
  • Inner: 5 * 6r = 30r
  • Last: 5 * 5 = 25

Adding these together, we get:

36r^2 + 30r + 30r + 25 = 36r^2 + 60r + 25

So, our expanded integrand is 36r^2 + 60r + 25. This makes the integration process much simpler because we can now integrate each term separately. Expanding the integrand is a crucial step because it transforms a potentially complex expression into a sum of simpler terms that are easier to integrate. Now that we have expanded the integrand, our integral looks like this: ∫(36r^2 + 60r + 25) dr from 1 to -1. Next, we'll find the antiderivative of this expanded expression.

Step 2: Finding the Antiderivative

Now that we've expanded the integrand, the next step is to find its antiderivative. The antiderivative is the reverse process of differentiation. We'll use the power rule for integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration. Since we're dealing with a definite integral, we don't need to worry about the constant of integration, as it will cancel out when we evaluate the integral at the limits.

Let's find the antiderivative of each term in our expanded integrand, 36r^2 + 60r + 25:

  • For 36r^2:
    • Applying the power rule, we increase the exponent by 1 (2 + 1 = 3) and divide by the new exponent:
    • ∫36r^2 dr = 36 * (r^3 / 3) = 12r^3
  • For 60r:
    • Here, the exponent of r is 1. Applying the power rule:
    • ∫60r dr = 60 * (r^2 / 2) = 30r^2
  • For 25:
    • This is a constant, and the integral of a constant is the constant times the variable:
    • ∫25 dr = 25r

Combining these results, the antiderivative of 36r^2 + 60r + 25 is 12r^3 + 30r^2 + 25r. So, we've successfully found the antiderivative! This means we're one step closer to solving the definite integral. The antiderivative represents the function whose derivative is our original integrand. Next, we'll evaluate this antiderivative at the limits of integration.

Step 3: Evaluating the Definite Integral

We've found the antiderivative, which is 12r^3 + 30r^2 + 25r. Now, we need to evaluate this antiderivative at the limits of integration, which are 1 and -1. This means we'll plug in the upper limit (1) into the antiderivative, then plug in the lower limit (-1), and subtract the result of the lower limit from the result of the upper limit. This process is based on the Fundamental Theorem of Calculus, which provides a straightforward method for evaluating definite integrals.

Let's start by evaluating the antiderivative at the upper limit, r = 1:

12(1)^3 + 30(1)^2 + 25(1) = 12 + 30 + 25 = 67

Now, let's evaluate the antiderivative at the lower limit, r = -1:

12(-1)^3 + 30(-1)^2 + 25(-1) = 12(-1) + 30(1) + 25(-1) = -12 + 30 - 25 = -7

Finally, we subtract the value at the lower limit from the value at the upper limit:

67 - (-7) = 67 + 7 = 74

So, the value of the definite integral ∫(6r+5)^2 dr from 1 to -1 is 74. We've successfully evaluated the integral by expanding the integrand, finding the antiderivative, and evaluating it at the limits of integration. This gives us the net signed area under the curve of the function (6r+5)^2 between r = 1 and r = -1.

Conclusion

Alright, guys! We've successfully evaluated the definite integral ∫(6r+5)^2 dr from 1 to -1. We started by expanding the integrand, then found its antiderivative using the power rule, and finally evaluated the antiderivative at the limits of integration. The final result is 74.

Remember, the key to solving these types of problems is to break them down into manageable steps. Expanding the integrand, finding the antiderivative, and evaluating at the limits are the core techniques you'll use for many definite integrals. Keep practicing, and you'll become a pro at solving these in no time! If you have any questions or want to tackle more integral problems, just let me know. Keep up the great work!