Definite Integrals: Properties And Evaluation Guide

Hey guys! Let's dive into the fascinating world of definite integrals. In this article, we're going to explore some key properties of definite integrals and then roll up our sleeves to evaluate a few juicy examples. Trust me, by the end of this, you'll be feeling like a definite integral pro!

Showing Integral Properties

Part (a)(i): Proving ∫₀² x² dx = ∫₀² t² dt

In this section, we're going to show that the definite integral of x² from 0 to 2 is equal to the definite integral of t² from 0 to 2. This might seem a bit obvious, but it's a crucial property of definite integrals that highlights how the variable of integration is just a placeholder. The key concept here is that the definite integral represents the area under the curve, and this area doesn't care what we call our variable!

To demonstrate this, let’s break it down. Remember, the definite integral of a function f(x) from a to b is defined as the limit of the Riemann sum as the number of subintervals approaches infinity. Or, more practically, it's the difference in the antiderivative evaluated at the upper and lower limits of integration.

So, let's first evaluate ∫₀² x² dx. The antiderivative of x² is (1/3)x³. Now, we need to evaluate this antiderivative at the limits of integration, 2 and 0, and subtract the results. That gives us [(1/3)(2)³] - [(1/3)(0)³] = (1/3)(8) - 0 = 8/3.

Now, let’s tackle ∫₀² t² dt. Guess what? The antiderivative of t² is (1/3)t³. We follow the same evaluation process: [(1/3)(2)³] - [(1/3)(0)³] = (1/3)(8) - 0 = 8/3. Look at that! Both integrals evaluate to the same value, 8/3. This beautifully illustrates that the variable of integration is indeed a dummy variable. It doesn’t matter if we use x or t; the definite integral remains the same because it's the area that matters, not the symbol we use to represent the variable along the x-axis.

This property is super useful because it allows us to switch variables in our integrals without changing the result, which can be handy in various integration techniques like substitution. So, the next time you see integrals with different variables but the same function and limits, remember, they're essentially the same thing!

Part (a)(ii): Proving ∫‡⁰ cos(x) dx = -∫₀‡ cos(x) dx

Okay, guys, let's dive into another cool property of definite integrals. This time, we're going to show that reversing the limits of integration changes the sign of the integral. In mathematical terms, we need to prove that ∫‡⁰ cos(x) dx = -∫₀‡ cos(x) dx. This property is all about understanding how direction matters when we're calculating the area under a curve.

To understand this, recall the fundamental definition of a definite integral. It's essentially the sum of infinitesimally small areas under the curve. When we integrate from a to b, we're moving along the x-axis in the positive direction. But when we switch the limits and integrate from b to a, we're moving in the opposite direction. This change in direction flips the sign of the integral.

Let’s break it down with our example. First, we find the antiderivative of cos(x), which is sin(x). Now, let's evaluate ∫‡⁰ cos(x) dx. We get sin(0) - sin(‡). We need the actual value of ‡, but without a specific value for ‡ we can't calculate sin(‡).

Next, let's evaluate -∫₀‡ cos(x) dx. This is equal to -[sin(‡) - sin(0)], which simplifies to -(sin(‡) - 0) = -sin(‡).

Now, comparing the two results, we see that sin(0) - sin(‡) = -sin(‡), which is the same as -[sin(‡) - sin(0)] = -sin(‡). This confirms our property: reversing the limits of integration changes the sign of the definite integral. In simpler terms, integrating backward gives you the negative of integrating forward. This property is particularly helpful when you need to manipulate integrals to fit certain forms or apply specific integration techniques. It's a neat trick to have up your sleeve!

Evaluating Definite Integrals

Part (b)(i): Evaluating ∫₀⁶ ⁴√(2x+4) dx

Alright, let's get our hands dirty and evaluate some definite integrals! Our first challenge is to find the value of ∫₀⁶ ⁴√(2x+4) dx. This integral involves a fourth root, which might look intimidating at first, but don't worry, we'll tackle it step by step. The key here is to use u-substitution, a powerful technique for simplifying integrals.

First, we need to identify a suitable substitution. Looking at the integral, the expression inside the fourth root, 2x+4, seems like a good candidate. So, let's set u = 2x + 4. Now, we need to find du, which is the derivative of u with respect to x, multiplied by dx. The derivative of 2x + 4 is simply 2, so du = 2 dx.

We also need to change our limits of integration to be in terms of u. When x = 0 (the lower limit), u = 2(0) + 4 = 4. When x = 6 (the upper limit), u = 2(6) + 4 = 16. So, our new limits of integration are 4 and 16.

Now, we can rewrite the integral in terms of u. We have ∫₀⁶ ⁴√(2x+4) dx. Since u = 2x + 4, we can rewrite ⁴√(2x+4) as u^(1/4). Also, we know du = 2 dx, so dx = (1/2) du. Plugging these into our integral, we get ∫₄¹⁶ u^(1/4) (1/2) du. We can pull the constant (1/2) out of the integral, giving us (1/2) ∫₄¹⁶ u^(1/4) du.

Now, we can find the antiderivative of u^(1/4). Using the power rule for integration, we add 1 to the exponent and divide by the new exponent: (u^(1/4 + 1)) / (1/4 + 1) = (u^(5/4)) / (5/4) = (4/5)u^(5/4). So, our antiderivative is (4/5)u^(5/4).

Finally, we need to evaluate this antiderivative at our limits of integration, 16 and 4, and subtract the results: (4/5)(16)^(5/4) - (4/5)(4)^(5/4). Let's break this down. 16^(5/4) is the fourth root of 16 raised to the power of 5. The fourth root of 16 is 2, so 16^(5/4) = 2⁵ = 32. Similarly, 4^(5/4) is the fourth root of 4 raised to the power of 5. The fourth root of 4 is √2, so 4^(5/4) = (√2)⁵ = 4√2.

Plugging these values back into our expression, we get (4/5)(32) - (4/5)(4√2) = 128/5 - (16√2)/5. So, the value of the definite integral ∫₀⁶ ⁴√(2x+4) dx is (128 - 16√2) / 5. Phew, that was a good one! We successfully used u-substitution to simplify the integral and found the exact value.

Part (b)(ii): Evaluating ∫₁² [4t + t(t²+1)³] dt

Okay, let's keep the ball rolling! Our next challenge is to evaluate ∫₁² [4t + t(t²+1)³] dt. This integral looks a bit more complex, but don't let it scare you. We'll use a combination of basic integration rules and u-substitution to solve it. The key is to break it down into manageable parts.

First, let's notice that we have two terms inside the integral: 4t and t(t²+1)³. We can integrate these separately and then add the results. Integrating 4t is straightforward: the antiderivative of 4t is 2t². Now, let's focus on the second term, t(t²+1)³. This looks like a perfect candidate for u-substitution.

Let's set u = t² + 1. Then, du = 2t dt. We have t dt in our integral, so we can rewrite this as (1/2) du = t dt. Now, we need to change our limits of integration. When t = 1 (the lower limit), u = 1² + 1 = 2. When t = 2 (the upper limit), u = 2² + 1 = 5. So, our new limits are 2 and 5.

We can rewrite the integral of the second term in terms of u: ∫ t(t²+1)³ dt becomes ∫ u³ (1/2) du. We can pull the (1/2) out, giving us (1/2) ∫ u³ du. The antiderivative of u³ is (1/4)u⁴. So, we have (1/2) * (1/4)u⁴ = (1/8)u⁴.

Now, we need to evaluate this antiderivative at our limits of integration, 5 and 2: (1/8)(5)⁴ - (1/8)(2)⁴ = (1/8)(625) - (1/8)(16) = 625/8 - 16/8 = 609/8. So, the integral of the second term is 609/8.

Now, let's go back to the first term, 4t. We found its antiderivative to be 2t². We need to evaluate this at our original limits of integration, 2 and 1: 2(2)² - 2(1)² = 2(4) - 2(1) = 8 - 2 = 6. So, the integral of the first term is 6.

Finally, we add the results of the two integrals: 6 + 609/8. To add these, we need a common denominator: 6 = 48/8. So, 48/8 + 609/8 = 657/8. Ta-da! The value of the definite integral ∫₁² [4t + t(t²+1)³] dt is 657/8. That was a bit more involved, but we conquered it using our integration skills!

Conclusion

So, guys, we've journeyed through the fascinating world of definite integrals, exploring some key properties and evaluating a couple of challenging examples. We've seen how the variable of integration is just a placeholder, how reversing the limits changes the sign, and how u-substitution can be a lifesaver when dealing with complex integrals. I hope you've gained a deeper understanding and appreciation for these powerful mathematical tools. Keep practicing, and you'll become a definite integral master in no time!