Approximate Integral Of 1/ln(x) From 3 To 5 Using Midpoint Rule

Hey guys! Today, we're diving into the fascinating world of numerical integration. Specifically, we're going to tackle the integral $\int_3^5 \frac{1}{\ln (x)} dx$ using the midpoint rule with three subdivisions. Don't worry if that sounds like a mouthful – we'll break it down step by step. Our goal is to approximate this integral to four decimal places. So, grab your calculators, and let's get started!

Understanding the Midpoint Rule

Before we jump into the calculations, let's quickly recap what the midpoint rule is all about. The midpoint rule is a numerical integration technique used to approximate the definite integral of a function. In essence, it's a way to find the area under a curve when we can't easily find the exact answer using traditional calculus methods. It's super handy for functions that don't have elementary antiderivatives, like our $\frac{1}{\ln (x)}$ friend here. The main idea behind the midpoint rule is to divide the interval of integration into smaller subintervals and then approximate the area under the curve in each subinterval by the area of a rectangle. The height of each rectangle is determined by the value of the function at the midpoint of the subinterval. So, instead of using the left or right endpoint like in other methods (like the left or right Riemann sums), we're using the middle ground, which often gives us a more accurate approximation. The formula for the midpoint rule is given by:

$$\int_a^b f(x) dx \approx \Delta x [f(x_1) + f(x_2) + ... + f(x_n)]$$

Where:

• $a$ and $b$ are the limits of integration.
• $n$ is the number of subintervals (subdivisions).
• $\Delta x = \frac{b-a}{n}$ is the width of each subinterval.
• $x_i$ is the midpoint of the $i$-th subinterval.

In simpler terms, we're chopping up the area under the curve into rectangles, finding the area of each rectangle using the midpoint height, and then adding those areas together. It's like tiling a slightly curvy floor with perfectly rectangular tiles – it won't be a perfect fit, but it'll be pretty close! We care about the midpoint rule because it often gives a better approximation compared to other basic numerical integration techniques like the trapezoidal rule or the left/right Riemann sums. This is because using the midpoint can help to balance out the errors that occur when approximating the area with rectangles. For a function that is concave up, the midpoint rule tends to overestimate the integral in each subinterval, while for a concave down function, it tends to underestimate. These errors tend to cancel each other out across the interval, leading to a more accurate overall approximation. That's why, in scenarios where accuracy is crucial but computational cost needs to be kept down, the midpoint rule is often the go-to method. So, with this fundamental concept in mind, we're now prepared to use the midpoint rule to approximate $\int_3^5 \frac{1}{\ln (x)} dx$.

Setting Up the Problem

Okay, let's apply this to our specific problem: approximating $\int_3^5 \frac{1}{\ln (x)} dx$ using the midpoint rule with three subdivisions ($n = 3$). Here's how we'll break it down:

1. Identify the limits of integration: We have $a = 3$ and $b = 5$.
2. Determine the number of subintervals: We're given $n = 3$.
3. Calculate the width of each subinterval ($\Delta x$): Using the formula $\Delta x = \frac{b-a}{n}$, we get $\Delta x = \frac{5-3}{3} = \frac{2}{3}$.
4. Find the endpoints of the subintervals: Starting from $a = 3$, we add $\Delta x$ successively to get the endpoints:
   • $x_0 = 3$
   • $x_1 = 3 + \frac{2}{3} = \frac{11}{3}$
   • $x_2 = \frac{11}{3} + \frac{2}{3} = \frac{13}{3}$
   • $x_3 = \frac{13}{3} + \frac{2}{3} = 5$
5. Determine the midpoints of each subinterval: The midpoint of an interval $[x_i, x_{i+1}]$ is simply the average of the endpoints:
   • $m_1 = \frac{3 + \frac{11}{3}}{2} = \frac{\frac{9}{3} + \frac{11}{3}}{2} = \frac{\frac{20}{3}}{2} = \frac{10}{3}$
   • $m_2 = \frac{\frac{11}{3} + \frac{13}{3}}{2} = \frac{\frac{24}{3}}{2} = \frac{8}{2} = 4$
   • $m_3 = \frac{\frac{13}{3} + 5}{2} = \frac{\frac{13}{3} + \frac{15}{3}}{2} = \frac{\frac{28}{3}}{2} = \frac{14}{3}$
6. Evaluate the function at each midpoint: Our function is $f(x) = \frac{1}{\ln (x)}$. So, we need to find $f(m_1)$, $f(m_2)$, and $f(m_3)$:
   • $f(m_1) = f(\frac{10}{3}) = \frac{1}{\ln (\frac{10}{3})}$
   • $f(m_2) = f(4) = \frac{1}{\ln (4)}$
   • $f(m_3) = f(\frac{14}{3}) = \frac{1}{\ln (\frac{14}{3})}$

So, now we've broken down our interval, found our midpoints, and have expressions for the function evaluated at these midpoints. We're in the home stretch now!

Calculating the Approximation

Alright, guys, now comes the crunching of the numbers! We're going to use the values we just calculated and plug them into the midpoint rule formula. Remember, the formula is:

$$\int_a^b f(x) dx \approx \Delta x [f(x_1) + f(x_2) + ... + f(x_n)]$$

In our case, this translates to:

$$\int_3^5 \frac{1}{\ln (x)} dx \approx \frac{2}{3} \left[ f(\frac{10}{3}) + f(4) + f(\frac{14}{3}) \right]$$

Now, let's substitute the values of $f(x)$ at the midpoints:

$$\int_3^5 \frac{1}{\ln (x)} dx \approx \frac{2}{3} \left[ \frac{1}{\ln (\frac{10}{3})} + \frac{1}{\ln (4)} + \frac{1}{\ln (\frac{14}{3})} \right]$$

This is where the calculator comes in handy. We need to compute the values of the natural logarithms and the reciprocals. Make sure your calculator is in radian mode (although it doesn't matter for logarithms, it's a good habit to check!). Calculating each term, we get:

• $\frac{1}{\ln (\frac{10}{3})} \approx \frac{1}{\ln (3.3333)} \approx \frac{1}{1.2040} \approx 0.8306$
• $\frac{1}{\ln (4)} \approx \frac{1}{1.3863} \approx 0.7213$
• $\frac{1}{\ln (\frac{14}{3})} \approx \frac{1}{\ln (4.6667)} \approx \frac{1}{1.5404} \approx 0.6492$

Plugging these approximations back into our formula:

$$\int_3^5 \frac{1}{\ln (x)} dx \approx \frac{2}{3} [0.8306 + 0.7213 + 0.6492]$$

$$\int_3^5 \frac{1}{\ln (x)} dx \approx \frac{2}{3} [2.2011]$$

$$\int_3^5 \frac{1}{\ln (x)} dx \approx 1.4674$$

Therefore, using the midpoint rule with three subdivisions, we approximate the integral $\int_3^5 \frac{1}{\ln (x)} dx$ to be approximately 1.4674. Remember, this is an approximation. The more subdivisions we use (i.e., the larger $n$ is), the more accurate our approximation will generally be. However, with just three subdivisions, we've gotten a pretty decent estimate! So, there you have it! We've successfully navigated the midpoint rule and approximated a tricky integral. High five!

Error Analysis (Optional, but insightful!)

Now, just for a bit of extra insight, let's briefly touch upon error analysis. How do we know how good our approximation is? There are a few ways to think about this, but one common approach is to consider the error bound for the midpoint rule. The error bound formula provides an upper limit on the error we can expect when using the midpoint rule. It's a guarantee that the true value of the integral is within a certain range of our approximation. The error bound for the midpoint rule is given by:

$$|Error| \leq \frac{K(b-a)^3}{24n^2}$$

Where:

• $K$ is the maximum value of the absolute value of the second derivative of $f(x)$ on the interval $[a, b]$.
• $a$ and $b$ are the limits of integration.
• $n$ is the number of subintervals.

Finding $K$ can be a bit tricky, as it involves finding the second derivative of our function and determining its maximum absolute value on the interval. For $f(x) = \frac{1}{\ln (x)}$, the second derivative is a bit messy:

$$f''(x) = \frac{1 + \ln (x)}{x^2(\ln (x))^3}$$

To find the maximum value of $|f''(x)|$ on $[3, 5]$, we'd typically need to analyze its critical points and endpoints. However, for the sake of brevity, let's assume (after some calculus elbow grease!) that the maximum value of $|f''(x)|$ on $[3, 5]$ is approximately $K \approx 0.1$. This is just an estimated value for illustration, and in a real-world scenario, you'd want to calculate it more precisely.

Now, we can plug our values into the error bound formula:

$$|Error| \leq \frac{0.1(5-3)^3}{24(3)^2} = \frac{0.1(8)}{24(9)} = \frac{0.8}{216} \approx 0.0037$$

This tells us that the error in our approximation is likely no more than 0.0037. This is a relatively small error bound, which indicates that our approximation of 1.4674 is reasonably accurate. Of course, the more subdivisions we use, the smaller this error bound would become. This error analysis reinforces the idea that numerical integration techniques, like the midpoint rule, give us powerful tools for estimating integrals, and understanding error bounds helps us gauge the reliability of those estimations. It's a blend of approximation and precision that makes numerical methods so crucial in mathematics and its applications!

Conclusion

So, guys, we've successfully approximated the integral $\int_3^5 \frac{1}{\ln (x)} dx$ using the midpoint rule with three subdivisions. We found an approximation of 1.4674. We also discussed the error analysis to give you a sense of how accurate our approximation is. The midpoint rule is a fantastic tool in our numerical integration toolkit, allowing us to tackle integrals that might otherwise be impossible to solve analytically. Remember, the key is to break the problem down into manageable steps: calculate $\Delta x$, find the midpoints, evaluate the function at those midpoints, and then plug everything into the formula. With a little practice, you'll be approximating integrals like a pro! Keep exploring, keep learning, and keep those calculators handy!