Evaluating The Integral Of 2e^(4x) + 3e^(2x) From -1 To 1

Hey math enthusiasts! Today, we're diving into a fun integral problem. We're going to evaluate the definite integral of the function 2e^(4x) + 3e^(2x) over the interval from -1 to 1. This is a classic calculus problem that combines exponential functions and the fundamental theorem of calculus. So, grab your calculators and let's get started!

Understanding the Integral

Before we jump into the calculations, let's make sure we understand what we're dealing with. The integral ∫[-1,1] (2e^(4x) + 3e^(2x)) dx represents the signed area between the curve of the function f(x) = 2e^(4x) + 3e^(2x) and the x-axis, from x = -1 to x = 1. Remember, areas above the x-axis are counted as positive, and areas below are counted as negative. Our main keyword here is integral, so keep that in mind as we move forward.

Breaking Down the Function

Our function, f(x) = 2e^(4x) + 3e^(2x), is a sum of two exponential terms. Exponential functions are those where the variable appears in the exponent, like e^(4x) or e^(2x). The constant 'e' is the base of the natural logarithm, approximately equal to 2.71828. When dealing with integrals, it's often helpful to break down complex functions into simpler parts, which is exactly what we'll do here. We have two main components: 2e^(4x) and 3e^(2x). Each of these will require a slightly different approach when we find their antiderivatives. Guys, understanding the components is crucial for solving the whole problem!

Why is this integral interesting?

This particular integral is a great example because it showcases several important concepts in calculus. Firstly, it requires us to understand the antiderivative of exponential functions. Secondly, it utilizes the fundamental theorem of calculus, which connects differentiation and integration. And thirdly, it involves evaluating a definite integral, which means we need to plug in the limits of integration and find the difference. This problem gives us a solid workout in these core calculus skills. So, let's dive deeper and see how we can solve it!

Finding the Antiderivative

The first step in evaluating a definite integral is to find the antiderivative of the function. The antiderivative, also known as the indefinite integral, is a function whose derivative is the original function. In other words, if F(x) is the antiderivative of f(x), then F'(x) = f(x). We'll find the antiderivative for each term in our function separately.

Antiderivative of 2e^(4x)

Let's start with 2e^(4x). We need to find a function whose derivative is 2e^(4x). Recall that the derivative of e^(kx) is ke^(kx), where k is a constant. So, the antiderivative of e^(kx) should be (1/k)e^(kx). Applying this to our term, we might guess that the antiderivative of 2e^(4x) is (2/4)e^(4x) = (1/2)e^(4x). To verify, let's differentiate (1/2)e^(4x): d/dx [(1/2)e^(4x)] = (1/2) * 4e^(4x) = 2e^(4x). Bingo! That's exactly what we wanted. Remember, folks, it's always a good idea to double-check your antiderivatives by differentiating them.

Antiderivative of 3e^(2x)

Now let's tackle 3e^(2x). Using the same logic as before, we look for a function whose derivative is 3e^(2x). The antiderivative of e^(2x) is (1/2)e^(2x). So, the antiderivative of 3e^(2x) should be (3/2)e^(2x). Let's verify: d/dx [(3/2)e^(2x)] = (3/2) * 2e^(2x) = 3e^(2x). Perfect! We've found the antiderivative for this term as well. It's like solving a puzzle, isn't it? Each piece fits perfectly into place.

Combining the Antiderivatives

Now that we have the antiderivatives of both terms, we can combine them to find the antiderivative of the entire function. The antiderivative of 2e^(4x) + 3e^(2x) is simply the sum of the individual antiderivatives: F(x) = (1/2)e^(4x) + (3/2)e^(2x). Don't forget, technically we should add a constant of integration, C, since the derivative of a constant is zero. However, for definite integrals, the constant will cancel out when we evaluate the limits, so we can ignore it for now. Our integral is starting to take shape, isn't it?

Applying the Fundamental Theorem of Calculus

The fundamental theorem of calculus is the bridge that connects differentiation and integration. It states that if F(x) is an antiderivative of f(x), then the definite integral of f(x) from a to b is given by: ∫[a,b] f(x) dx = F(b) - F(a). In simpler terms, to evaluate a definite integral, we find the antiderivative, plug in the upper limit of integration, plug in the lower limit of integration, and subtract the results. This is the heart of solving definite integrals, guys! This theorem is the key to unlocking the solution.

Evaluating at the Upper Limit (x = 1)

Our upper limit of integration is 1. So, we need to evaluate our antiderivative, F(x) = (1/2)e^(4x) + (3/2)e^(2x), at x = 1: F(1) = (1/2)e^(41) + (3/2)e^(21) = (1/2)e^4 + (3/2)e^2. This gives us a numerical value that represents the accumulated area under the curve up to x = 1. Remember, 'e' is just a number, so e^4 and e^2 are also just numbers. We're getting closer to the final answer!

Evaluating at the Lower Limit (x = -1)

Next, we need to evaluate our antiderivative at the lower limit of integration, which is -1: F(-1) = (1/2)e^(4*(-1)) + (3/2)e^(2*(-1)) = (1/2)e^(-4) + (3/2)e^(-2). This value represents the accumulated area under the curve up to x = -1. Notice the negative exponents! Remember that e^(-x) is the same as 1/e^x. So, e^(-4) = 1/e^4 and e^(-2) = 1/e^2. Now we have both F(1) and F(-1). We're almost there!

Subtracting to Find the Definite Integral

Finally, we subtract F(-1) from F(1) to find the definite integral: ∫[-1,1] (2e^(4x) + 3e^(2x)) dx = F(1) - F(-1) = [(1/2)e^4 + (3/2)e^2] - [(1/2)e^(-4) + (3/2)e^(-2)]. This subtraction gives us the net signed area between the curve and the x-axis over the interval from -1 to 1. We can simplify this expression a bit: (1/2)(e^4 - e^(-4)) + (3/2)(e^2 - e^(-2)). Guys, this is the exact value of the definite integral!

Calculating the Numerical Value

While the expression (1/2)(e^4 - e^(-4)) + (3/2)(e^2 - e^(-2)) is the exact answer, it's often helpful to get a numerical approximation. We can use a calculator to find the values of e^4, e^(-4), e^2, and e^(-2), and then plug them into our expression. Using a calculator, we find that: e^4 ≈ 54.598, e^(-4) ≈ 0.018, e^2 ≈ 7.389, e^(-2) ≈ 0.135. Plugging these values into our expression, we get: (1/2)(54.598 - 0.018) + (3/2)(7.389 - 0.135) ≈ (1/2)(54.58) + (3/2)(7.254) ≈ 27.29 + 10.881 ≈ 38.171. So, the numerical value of the definite integral is approximately 38.171. This gives us a sense of the magnitude of the area under the curve.

Conclusion

And there you have it! We've successfully evaluated the definite integral of 2e^(4x) + 3e^(2x) from -1 to 1. We found the antiderivative, applied the fundamental theorem of calculus, and calculated the numerical value. This problem is a great illustration of how powerful calculus can be in solving real-world problems. Remember, the key to mastering integrals is practice, practice, practice! So, keep exploring, keep learning, and keep having fun with math! Keep shining, everyone!