Integral Of E^x - E^-x From 0 To Ln(9)

by Andrew McMorgan 39 views

Hey math enthusiasts, let's dive into another cool calculus problem! Today, we're tackling the evaluation of a definite integral: ∫₀^ln(9) (eˣ - e⁻ˣ) dx. This problem involves exponential functions, specifically e raised to the power of x and e raised to the power of negative x, integrated over a specific interval from 0 to the natural logarithm of 9. It's a great way to practice your integration skills and understand how to handle definite integrals with logarithmic limits. So grab your calculators (or just your brains!), and let's break this down step-by-step. We'll make sure to express our numbers in their exact forms, using fractions and symbolic notation wherever needed, so no approximations here, guys!

Understanding the Integrand: eˣ - e⁻ˣ

First off, let's get friendly with our integrand, which is (eˣ - e⁻ˣ). This expression is actually quite significant in calculus and has a special name: it's the hyperbolic sine function, often denoted as sinh(x). So, the integral we're evaluating is essentially ∫₀^ln(9) sinh(x) dx. Knowing this can sometimes simplify how you think about the problem, but even without recognizing it, the integration process is straightforward. The function eˣ is its own derivative and integral, which is super handy. The function e⁻ˣ, on the other hand, has a derivative of -e⁻ˣ and an integral of -e⁻ˣ. So, when we integrate (eˣ - e⁻ˣ), we'll integrate each term separately. This is a fundamental property of integrals: the integral of a sum or difference is the sum or difference of the integrals. This makes tackling more complex expressions much more manageable. We are looking for the antiderivative of (eˣ - e⁻ˣ). The antiderivative of eˣ is simply eˣ. The antiderivative of -e⁻ˣ is e⁻ˣ (because the derivative of e⁻ˣ is -e⁻ˣ, so the integral of -e⁻ˣ is e⁻ˣ). Therefore, the antiderivative of (eˣ - e⁻ˣ) is (eˣ + e⁻ˣ). This combination (eˣ + e⁻ˣ) is also a famous hyperbolic function, the hyperbolic cosine, or cosh(x). So, we've found our antiderivative, and it's cosh(x). Keep this in mind as we move on to applying the limits of integration.

Applying the Fundamental Theorem of Calculus

Now that we have the antiderivative, (eˣ + e⁻ˣ), we can apply the Fundamental Theorem of Calculus. This theorem tells us that to evaluate a definite integral from a to b of a function f(x), we find the antiderivative F(x) and then calculate F(b) - F(a). In our case, the antiderivative is F(x) = eˣ + e⁻ˣ, the upper limit of integration is b = ln(9), and the lower limit is a = 0. So, we need to compute [eˣ + e⁻ˣ]₀^ln(9). This means we'll substitute ln(9) for x in our antiderivative, and then subtract the result of substituting 0 for x.

Let's evaluate the antiderivative at the upper limit, x = ln(9):

  • e^(ln(9)): This is the first part. Remember that the exponential function eˣ and the natural logarithm function ln(x) are inverse functions. This means that e^(ln(y)) = y for any positive y. So, e^(ln(9)) = 9. How cool is that? It simplifies our expression considerably.
  • e^(-ln(9)): Now for the second part. We can use the properties of logarithms and exponents here. Recall that -ln(y) = ln(y⁻¹). Therefore, -ln(9) = ln(9⁻¹), which is ln(1/9). So, e^(-ln(9)) = e^(ln(1/9)). Applying the inverse function property again, we get e^(ln(1/9)) = 1/9. Alternatively, you can think of it as e^(-ln(9)) = 1 / e^(ln(9)) = 1/9. This property is super useful when dealing with negative exponents and logarithms.

So, when x = ln(9), our antiderivative eˣ + e⁻ˣ becomes 9 + 1/9. To add these, we find a common denominator, which is 9. So, 9 + 1/9 = 81/9 + 1/9 = 82/9. This is the value of our antiderivative at the upper limit.

Now, let's evaluate the antiderivative at the lower limit, x = 0:

  • e⁰: Any non-zero number raised to the power of 0 is 1. So, e⁰ = 1.
  • e⁻⁰: This is the same as e⁰, which is also 1.

So, when x = 0, our antiderivative eˣ + e⁻ˣ becomes 1 + 1 = 2. This is the value of our antiderivative at the lower limit.

Finally, we subtract the value at the lower limit from the value at the upper limit: F(b) - F(a) = (82/9) - 2. To subtract these, we again find a common denominator, which is 9. So, 2 = 18/9. Therefore, 82/9 - 18/9 = (82 - 18) / 9 = 64/9. And there you have it, the exact value of our definite integral!

Final Result and Verification

The evaluation of the integral ∫₀^ln(9) (eˣ - e⁻ˣ) dx results in 64/9. We arrived at this by first finding the antiderivative of (eˣ - e⁻ˣ), which is (eˣ + e⁻ˣ). Then, we applied the Fundamental Theorem of Calculus by evaluating this antiderivative at the upper limit ln(9) and the lower limit 0, and subtracting the latter from the former. The key steps involved using the inverse properties of the exponential and natural logarithm functions, specifically e^(ln(y)) = y and e^(-ln(y)) = 1/y, and remembering that any non-zero number raised to the power of 0 is 1. The arithmetic led us to (9 + 1/9) - (1 + 1) = 82/9 - 2 = 64/9. It's always a good idea to double-check your work, especially with the properties of exponents and logarithms. Did we correctly integrate e⁻ˣ? Yes, the integral of e⁻ˣ is -e⁻ˣ, but our original integrand was (eˣ - e⁻ˣ). So the integral of eˣ is eˣ and the integral of -e⁻ˣ is e⁻ˣ. Therefore, the antiderivative is indeed (eˣ + e⁻ˣ). The evaluation at the limits seems correct too. The use of exact forms ensures precision, which is crucial in mathematics. This result, 64/9, is the precise area under the curve of the function (eˣ - e⁻ˣ) between x = 0 and x = ln(9). Pretty neat, right? Keep practicing these types of integrals, and you'll become a calculus whiz in no time!