Area Calculation: 6x+5y=30, X-axis, X=-1, And X=3
Alright, let's dive into a fun problem! We need to find the area of the region that's trapped between the line 6x + 5y = 30, the x-axis, and the vertical lines x = -1 and x = 3. And just to make it clear, we only care about the area that sits above the x-axis. Let's break this down step by step so we can nail this problem.
Step 1: Understanding the Line
First things first, we need to get a good handle on our line, 6x + 5y = 30. It's not in the friendliest format right now, so let's rearrange it to the slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. This will make it much easier to visualize and work with.
So, let's isolate y:
5y = -6x + 30
y = (-6/5)x + 6
Okay, now we can see that our line has a slope of -6/5 and a y-intercept of 6. This means the line slopes downwards as we move from left to right, and it crosses the y-axis at the point (0, 6). Knowing this helps us understand how the line behaves in the coordinate plane.
Step 2: Visualizing the Region
Now that we know what our line looks like, let's think about the region we're interested in. We have the x-axis as one boundary, and the vertical lines x = -1 and x = 3 chop off the region on the left and right sides. The line y = (-6/5)x + 6 forms the upper boundary of our region, but only where it's above the x-axis. Imagine drawing this on a graph to get a clear picture. This will help prevent errors. Remember only care about the part of the region that lies above the x-axis, since that's what the problem specifies.
Step 3: Setting Up the Integrals
To find the area of this region, we're going to use integration. The basic idea is to integrate the function that defines the upper boundary of our region (which is our line) with respect to x, between the limits of integration given by the vertical lines x = -1 and x = 3. However, we need to make sure our line is above the x-axis within this interval. If it dips below the x-axis, we'll need to adjust our approach a bit.
Let's find where the line intersects the x-axis by setting y = 0:
0 = (-6/5)x + 6
(6/5)x = 6
x = 5
So, the line crosses the x-axis at x = 5. Since our region is bounded by x = -1 and x = 3, the line y = (-6/5)x + 6 is always above the x-axis within our interval. That simplifies things quite a bit!
Now we can set up our integral. The area A is given by:
A = ∫[-1 to 3] ((-6/5)x + 6) dx
Step 4: Evaluating the Integrals
Alright, let's calculate that integral. We'll integrate the function (-6/5)x + 6 with respect to x, and then evaluate the result at our limits of integration, x = -1 and x = 3.
The integral of (-6/5)x is (-3/5)x^2, and the integral of 6 is 6x. So, we have:
∫ ((-6/5)x + 6) dx = (-3/5)x^2 + 6x + C
Now, we evaluate this at x = 3 and x = -1 and subtract to find the definite integral:
[(-3/5)(3)^2 + 6(3)] - [(-3/5)(-1)^2 + 6(-1)]
[(-3/5)(9) + 18] - [(-3/5)(1) - 6]
[-27/5 + 18] - [-3/5 - 6]
[-27/5 + 90/5] - [-3/5 - 30/5]
[63/5] - [-33/5]
63/5 + 33/5
96/5
So, the area A is 96/5 square units, or 19.2 square units.
Step 5: Putting It All Together
So, after all that calculating, we've found that the area of the region bounded by the line 6x + 5y = 30, the x-axis, and the lines x = -1 and x = 3 is 96/5 square units, which is 19.2 square units. Awesome job! You've successfully navigated through the problem by understanding the geometry, setting up the integral, and grinding through the calculations. Remember to double-check your work to avoid any simple arithmetic errors, and keep practicing to build your confidence with these types of problems.
Key takeaways
- When finding areas bounded by curves, it's crucial to understand the geometry of the region. Sketching the graph helps a lot.
- Setting up the integral correctly is key. Make sure you have the correct limits of integration and the correct function to integrate.
- Take your time with the calculations to avoid mistakes. Double-checking your work is always a good idea.
Practice Problems
Problem 1
Find the area of the region bounded by the line y = 2x + 3, the x-axis, and the lines x = 0 and x = 2. This will test your understanding of basic linear functions and integration. Remember to set up the integral correctly and evaluate it carefully!
Problem 2
Determine the area enclosed by the curve y = x^2 - 4x + 3 and the x-axis between its x-intercepts. This problem involves finding the roots of a quadratic equation and then integrating the function between those roots. Don't forget to consider whether the function is above or below the x-axis in the interval!
Problem 3
Calculate the area of the region bounded by the curves y = x^2 and y = 4x. This one requires finding the points of intersection between the two curves and then integrating the difference between the functions over the interval defined by these points. Make sure to identify which function is on top and which is on the bottom.
Conclusion
Calculating the area of a region bounded by a line and the x-axis involves understanding the line's equation, visualizing the region, setting up the integral, and evaluating it correctly. It's a process that combines algebra, geometry, and calculus, so make sure you're comfortable with all three! Keep practicing, and you'll become a pro at these problems in no time!
