Average Value Of E^(2x) On [0, Ln 5]

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of calculus, specifically tackling a problem that might seem a bit intimidating at first glance: finding the average value of a function over a given interval. If you're into mathematics, this is right up your alley! We'll be working with the function $f(x) = e^{2x}$ and the interval [0, oldsymbol{\ ext{ln } 5}]. Don't worry if exponential functions and logarithms give you a slight pause; by the end of this article, you'll have a solid grasp on how to find the average value, and you'll be able to impress your friends with your newfound calculus prowess. So, grab a coffee, get comfortable, and let's get started on this mathematical adventure!

Understanding the Average Value of a Function

Before we jump into the nitty-gritty of solving this problem, let's make sure we're all on the same page about what the average value of a function actually means. In calculus, the average value of a function $f(x)$ over an interval $[a, b]$ is essentially the height of a rectangle that has the same area as the region under the curve of $f(x)$ from $a$ to $b$. Think of it this way: if you were to smooth out all the ups and downs of the function's graph over that interval, the average value would be the constant height it would have. Mathematically, this concept is defined by the following formula:

$$f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx$$

Here, $f_{\text{avg}}$ represents the average value, $a$ and $b$ are the endpoints of our interval, and $\int_{a}^{b} f(x) \, dx$ is the definite integral of the function $f(x)$ from $a$ to $b$. The $\frac{1}{b-a}$ term normalizes the integral, ensuring we get an average value over the specified length of the interval. This formula is a cornerstone of integral calculus and has applications in various fields, from physics and engineering to economics and statistics. It allows us to represent the overall behavior of a function across an interval with a single, representative value. So, when we're asked to find the average value of $f(x)=e^{2x}$ over [0, oldsymbol{\ ext{ln } 5}], we're essentially looking for that single height that would give us the same area under the curve as the original, more complex function. It's a powerful tool for simplifying complex data and understanding trends. We'll be using this formula as our guide throughout this problem. Ready to apply it?

Applying the Formula to Our Function

Alright, guys, let's put the formula into action! We are given the function $f(x) = e^{2x}$ and the interval [0, oldsymbol{\ ext{ln } 5}]. According to our formula for the average value, we need to calculate:

$$f_{\text{avg}} = \frac{1}{(\ln 5) - 0} \int_{0}^{\ln 5} e^{2x} \, dx$$

First things first, let's simplify the denominator: (oldsymbol{\ ext{ln } 5}) - 0 = oldsymbol{\ ext{ln } 5}. So, our expression becomes:

$$f_{\text{avg}} = \frac{1}{\ln 5} \int_{0}^{\ln 5} e^{2x} \, dx$$

Now, the core of our problem lies in evaluating the definite integral $\int_{0}^{\ln 5} e^{2x} \, dx$. To do this, we first need to find the antiderivative of $e^{2x}$. Remember your integration rules? The integral of $e^{kx}$ is $\frac{1}{k} e^{kx} + C$. In our case, $k=2$. So, the antiderivative of $e^{2x}$ is $\frac{1}{2} e^{2x}$.

Now we apply the Fundamental Theorem of Calculus. We need to evaluate the antiderivative at the upper limit (oldsymbol{\ ext{ln } 5}) and subtract its value at the lower limit (0):

$$\int_{0}^{\ln 5} e^{2x} \, dx = \left[ \frac{1}{2} e^{2x} \right]_{0}^{\ln 5} = \frac{1}{2} e^{2(\ln 5)} - \frac{1}{2} e^{2(0)}$$

Let's simplify this further. Recall that $e^{2(\ln 5)} = e^{\ln (5^2)} = 5^2 = 25$. And $e^{2(0)} = e^0 = 1$. So, the value of the definite integral is:

$$\frac{1}{2}(25) - \frac{1}{2}(1) = \frac{25}{2} - \frac{1}{2} = \frac{24}{2} = 12$$

Awesome! We've successfully evaluated the integral. The value of the definite integral is 12. This means the area under the curve of $f(x) = e^{2x}$ from $x=0$ to x=oldsymbol{\ ext{ln } 5} is exactly 12 square units. Pretty neat, right? This integral represents the accumulation of the function's values over the interval. It's a fundamental step in understanding how functions behave and what their overall impact is. Now, all that's left is to plug this value back into our average value formula.

Calculating the Final Average Value

We've done the heavy lifting, guys! We found that the definite integral $\int_{0}^{\ln 5} e^{2x} \, dx$ equals 12. Now, we just need to substitute this back into our average value formula:

$$f_{\text{avg}} = \frac{1}{\ln 5} \int_{0}^{\ln 5} e^{2x} \, dx$$

$$f_{\text{avg}} = \frac{1}{\ln 5} (12)$$

$$f_{\text{avg}} = \frac{12}{\ln 5}$$

And there you have it! The average value of the function $f(x) = e^{2x}$ over the interval [0, oldsymbol{\ ext{ln } 5}] is $\boldsymbol{\rac{12}{\ln 5}}$. Looking back at the options provided, this matches option (c).

This result tells us that if we were to draw a horizontal line at the height of $\boldsymbol{\rac{12}{\ln 5}}$ on our graph, the area between this line and the x-axis over the interval [0, oldsymbol{\ ext{ln } 5}] would be exactly the same as the area under the curve of $f(x)=e^{2x}$ over the same interval. It's a way to find a single, representative value for a function's behavior across a range. This concept is super useful in many real-world applications where we need to understand the average rate of change or average magnitude of a varying quantity. For instance, if $e^{2x}$ represented the velocity of an object, $\boldsymbol{\rac{12}{\ln 5}}$ would be its average velocity over the given time interval. It simplifies complex motion into a single, understandable figure.

Why This Matters: Applications of Average Value

So, why should you care about finding the average value of a function? Well, this concept, while rooted in abstract mathematics, has some seriously cool real-world applications. Think about it: many natural phenomena and engineered systems involve quantities that change over time or space. To understand these systems better, we often need to find an average measure of these changing quantities. For example, in physics, if you have a variable force acting on an object over a distance, you can calculate the average force applied. Similarly, if the temperature of a room fluctuates throughout the day, you can find the average temperature. In engineering, this concept is used to calculate things like the average stress on a material or the average flow rate in a pipe. In economics, it can help in understanding average prices or average income over a period.

The formula for the average value of a function is a direct consequence of the Mean Value Theorem for Integrals, which itself is a significant theorem in calculus. It guarantees that for any continuous function on a closed interval, there exists at least one point within that interval where the function's value is equal to its average value. This means our calculated average, $\boldsymbol{\rac{12}{\ln 5}}$, is not just a theoretical number; it's an actual value that the function $f(x) = e^{2x}$ takes on at some point within the interval [0, oldsymbol{\ ext{ln } 5}]. This connection makes the average value a concrete and meaningful representation of the function's behavior. It provides a single value that summarizes the function's performance over the entire interval, making it easier to compare different functions or different intervals. It’s a powerful tool for simplifying complex information and making informed decisions based on trends and overall performance. So, the next time you see a calculation for an average value, remember it’s more than just math – it’s a tool for understanding the world around us!

Conclusion

And that, my friends, is how you find the average value of a function! We took the function $f(x) = e^{2x}$ and the interval [0, oldsymbol{\ ext{ln } 5}], applied the average value formula, evaluated the definite integral using the Fundamental Theorem of Calculus, and arrived at our answer: $\boldsymbol{\rac{12}{\ln 5}}$. We saw that this value represents a constant height that would yield the same area under the curve as our original function over the given interval. This concept is fundamental in calculus and has widespread applications across various scientific and economic fields. Keep practicing these types of problems, and you'll become a calculus whiz in no time! Remember, understanding these core concepts is key to unlocking more complex mathematical ideas and appreciating the elegance of calculus. Don't shy away from the challenges; embrace them as opportunities to learn and grow. Until next time, happy calculating!