Calculating Areas: Cosines, Sines & Linear Equations

Hey guys! Let's dive into some calculus and figure out how to find the area of regions bounded by curves. This is a super important concept, so pay attention! We're gonna break down each part step-by-step, making sure we cover everything from the basics to the nitty-gritty details. Ready to get started?

Part a) Analyzing $y = cos x$, $y = -x$, $x = 0$, x = rac{\pi}{2} for Area Calculation

Alright, first up, let's tackle part (a). Here, we're dealing with the curves $y = cos x$, $y = -x$, the vertical lines $x = 0$, and $x = \frac{\pi}{2}$. Our mission? To calculate the area enclosed by these bad boys. The key here, and what you’ll always need to do, is visualize the situation. A simple sketch can work wonders! It helps you understand which function is 'on top' and which is 'on the bottom' within the given interval. This is crucial because the area is calculated by integrating the difference between the functions. Let's do a quick rundown of how to approach this. First, sketch the graphs. Plot $y = cos x$ and $y = -x$ on the same coordinate plane. The $x = 0$ and $x = \frac{\pi}{2}$ lines will define our boundaries along the x-axis. Notice that within the interval $[0, \frac{\pi}{2}]$, the cosine function starts at 1, decreases, and intersects with $y=-x$. We know that $cos x$ is greater than $-x$ at least at the beginning of the interval, starting at $x = 0$. So we know we will have to find intersection point to know what order of function we need to use in the integral. That intersection point is between $0$ and $\frac{\pi}{2}$.

Now, how to compute it? The area between two curves, $f(x)$ and $g(x)$, from $x = a$ to $x = b$, is given by the integral of the absolute difference of the functions. In other words, $\int_{a}^{b} |f(x) - g(x)| dx$. This basically means, you subtract one function from the other, and integrate over the interval. So the area we want to calculate is the integral from 0 to $\frac{\pi}{2}$ of the difference between the two functions. In this case, since we want the area, we will need to calculate the intersection point, $x_0$, and make two integrals, one from $0$ to $x_0$, and another from $x_0$ to $\frac{\pi}{2}$. And in each of the intervals, you will have to determine the order of the functions. Let's suppose $x_0$ is the intersection point, now the formula is: $ \int_{0}^{x_0} (cos x - (-x)) dx + \int_{x_0}^{\frac{\pi}{2}} (-x - cos x) dx$. Compute the antiderivatives of each function. For $cos x$, the antiderivative is $sin x$. For $-x$, the antiderivative is $-\frac{x^2}{2}$. Evaluate the antiderivatives at the boundaries of each interval, and subtract the results to find the definite integral for each section of area. Add the results. So, after finding the intersection point, calculating the integral, and doing all the calculations, we will have our area. Remember, visualization and careful calculation are your best friends here. You got this!

Part б) Calculating the Area for $y = sin 2x$, $y = x - \frac{\pi}{2}$, $x = 0$

On to part (б)! Here, we've got $y = sin 2x$, $y = x - \frac{\pi}{2}$, and $x = 0$. This one's a little different, but the same principles apply. First, let's sketch the graphs again. The curve $y = sin 2x$ oscillates, which means it goes up and down, and the line $y = x - \frac{\pi}{2}$ is a straight line. The line $x = 0$ is the y-axis, and acts as a boundary. So, we're finding the area between the sine curve and the line, starting at the y-axis, and extending until the intersection point between the curves. What's the strategy? We have to compute the integral of the difference between the two functions. And we'll use the same formula as before, it is always the integral of the absolute difference. The difference is the order of the functions. You will have to determine which function is on top, and which is on the bottom. Start by finding the intersection point(s) of the curves. To do this, set the two functions equal to each other, $sin 2x = x - \frac{\pi}{2}$, and solve for $x$. This might require some numerical methods, since it's not a straightforward algebraic solution. Once you know the intersection point, call it $x_1$, and determine which function is on top within the interval $[0, x_1]$. Now we know our bounds. To calculate the area, integrate the absolute difference of the functions, from $0$ to $x_1$. So the area will be: $ \int_{0}^{x_1} (sin 2x - (x - \frac{\pi}{2})) dx$. The antiderivative of $sin 2x$ is $-\frac{1}{2}cos 2x$, and the antiderivative of $x - \frac{\pi}{2}$ is $\frac{x^2}{2} - \frac{\pi}{2}x$. Substitute the upper and lower bounds into the antiderivatives, and you have your area! Remember to be careful with the signs and that a sketch can guide you to find the correct order of the functions. Make sure you're comfortable with both integration and basic algebra. This is where practice really pays off. Keep at it, and you’ll master this in no time!

Part в) Solving the Area Problem with $y = sin x$, $y = -x$, $x = 0$, $x = \frac{\pi}{2}$!

Let's get cracking on part (в)! We've got $y = sin x$, $y = -x$, $x = 0$, and $x = \frac{\pi}{2}$. This problem is similar to part (a) in its structure, but with different functions. Let's make sure we understand the functions and their properties. We will first need to find the intersection point between $sin x$ and $-x$. To find the intersection points, set the two functions equal to each other: $sin x = -x$. One solution is at $x = 0$. Another one is between $\frac{\pi}{2}$. This means we will have to calculate two integrals. Let's analyze the problem. We want to find the area bounded by these curves. We know that the limits for the $x$ values are from 0 to $\frac{\pi}{2}$. The critical step is to determine which function is 'above' the other in the interval. The function that is 'above' will be the one that is the greatest, or the function that has a higher value than the other. Start by sketching $y = sin x$ and $y = -x$ on a graph. The sine function oscillates, going up and down, and the function $y = -x$ is a straight line sloping downwards. At $x = 0$, we have that $sin(0) = 0$ and $-0 = 0$. They intersect. The intersection point is $0$. Therefore, in the interval $[0, \frac{\pi}{2}]$, we have to calculate: $\int_{0}^{\frac{\pi}{2}} (sin x - (-x)) dx$. Now, we need to calculate the integral. The antiderivative of $sin x$ is $-cos x$, and the antiderivative of $-x$ is $-\frac{x^2}{2}$. So now we evaluate the integral: $ \int_{0}^{\frac{\pi}{2}} (sin x - (-x)) dx = -cos x + \frac{x^2}{2} \Big|_0^{\frac{\pi}{2}} = (-cos(\frac{\pi}{2}) + \frac{(\frac{\pi}{2})^2}{2}) - (-cos(0) + \frac{0^2}{2}) $. Now it is only a matter of calculating. Evaluate the antiderivative at the limits of integration. You're essentially finding the net area, so positive and negative areas will be accounted for. The calculation is mostly the same as before, but with different functions, so remember to make sure the order of functions are correct. You are getting better at this!

Part г) Calculating the Area for $y = cos \frac{x}{2}$, $y = x - \pi$, $x = 0$, $x = \pi$

Alright, let's finish with part (г)! This one gives us $y = cos \frac{x}{2}$, $y = x - \pi$, $x = 0$, and $x = \pi$. This time, our interval on the x-axis is from 0 to π. The curve $y = cos \frac{x}{2}$ is a cosine function that has a longer period. The line $y = x - \pi$ is a straight line. Follow the same strategy. First, sketch the graphs to visualize the problem. Determine the intersection points, if any, within the given interval. To determine the intersection points, set the functions equal: $cos \frac{x}{2} = x - \pi$. Solving this might be tricky without numerical methods, but you can estimate from your graph or use a calculator. You will find that the curves intersect. Now, determine which function is on top within the interval $[0, \pi]$. Based on your sketch, the cosine curve appears to be above the line. To confirm, you can test values within the interval. Once you know which function is 'on top', set up your integral. The area will be $ \int_0}^{\pi} (cos \frac{x}{2} - (x - \pi)) dx$. Now, let's calculate the integral. The antiderivative of $cos \frac{x}{2}$ is $2sin \frac{x}{2}$, and the antiderivative of $x - \pi$ is $\frac{x^2}{2} - \pi x$. Now calculate the integral, by applying the boundaries $ 2sin \frac{x{2} - \frac{x^2}{2} + \pi x \Big|_0^{\pi}$. Apply the boundaries, do all the calculations and you will have your area! Remember, the core process remains the same: sketch, determine the order, and integrate! With consistent practice, these problems will become second nature to you. Keep up the awesome work!