R is the region below the curve y=x and above the z-axis from x=0 to x=b, where b is a positive constant. S is the region below the curve y=cos(x) and above the x-axis from x=0 to x=b. For what value of b is the area of R equal to the area of S?
I know the answer is 1.404, but I need an explanation as to why. Please and thank you.
The area of R is equal to the area of S when the integral of y=x from 0 to b is equal to the integral of y=cos(x) from 0 to b. To solve for b, we can set the two integrals equal to each other and solve for b.
Integral of y=x from 0 to b = Integral of y=cos(x) from 0 to b
∫x dx from 0 to b = ∫cos(x) dx from 0 to b
(1/2)x^2 from 0 to b = sin(x) from 0 to b
(1/2)b^2 = sin(b) - sin(0)
(1/2)b^2 = sin(b)
b^2 = 2sin(b)
b = sqrt(2sin(b))
We can solve for b numerically using a calculator or computer program. The answer is b = 1.404.
Ah, the battle of the curves! Let's dive into it with a splash of humor to make it more enjoyable.
First, let's find the area of region R, where y = x. To calculate the area under this curve, we need to integrate it from x = 0 to x = b. So, the area of R is given by the integral of x dx from 0 to b.
Now, let's move on to region S, where y = cos(x). Similar to R, we want to find the area under the curve from x = 0 to x = b. So, the area of S is given by the integral of cos(x) dx from 0 to b.
To find the value of b where the areas of R and S are equal, we need to set up an equation. So we have:
∫(x dx) from 0 to b = ∫(cos(x) dx) from 0 to b.
Integrating both sides, we get:
(b^2)/2 - 0 = sin(b) - sin(0).
Now, let's simplify:
(b^2)/2 = sin(b).
Using a bit of algebraic magic, we rearrange the equation to isolate b:
b^2 = 2sin(b).
To find the exact value of b where the areas are equal, we consult our trusty calculators or mathematical tools. After crunching the numbers, we discover that b ≈ 1.404.
And that's the punchline of the story! The value of b where the areas of R and S are equal is approximately 1.404.
Keep in mind, though, that this explanation might not be the most rigorous or formal. But hey, math with a touch of humor never hurts!
To find the value of b for which the area of region R is equal to the area of region S, we need to set up and solve the following integral equations:
Area of R: ∫[0,b] (x) dx
Area of S: ∫[0,b] (cos(x)) dx
Let's calculate the area of R first:
∫[0,b] (x) dx = 1/2 * x^2 |[0,b] = 1/2 * b^2
Now let's calculate the area of S:
∫[0,b] (cos(x)) dx = sin(x) |[0,b] = sin(b)
We want to find the value of b such that the area of R is equal to the area of S:
1/2 * b^2 = sin(b)
To solve this equation, we can use numerical methods or an approximate method like the Newton-Raphson method. Using numerical methods, we find that the value of b where the areas are equal is approximately 1.404.
Thus, the area of R is equal to the area of S when b ≈ 1.404.
To find the value of b for which the areas of regions R and S are equal, we need to calculate the areas of both regions and set them equal to each other.
Let's begin with region R.
The curve y = x is a straight line passing through the origin (0,0) with a slope of 1. We need to find the x-coordinate where this line intersects the x-axis. Since the line passes through the origin, the x-intercept is simply 0. Therefore, the region R lies from x = 0 to x = b.
To find the area of region R, we integrate the equation of the curve y = x with respect to x over the interval [0, b]:
Area of R = ∫[0, b] x dx.
Integrating x with respect to x gives us the anti-derivative x^2/2. Evaluating this anti-derivative at the upper limit b and lower limit 0, we have:
Area of R = [x^2/2] from 0 to b = (b^2/2) - (0^2/2) = b^2/2.
Now let's consider region S.
The curve y = cos(x) is a periodic function oscillating between -1 and 1 with respect to the x-axis. We need to find the x-coordinate(s) where this curve intersects the x-axis within the interval [0, b].
To find the area of region S, we integrate the equation of the curve y = cos(x) with respect to x over the interval [0, b]:
Area of S = ∫[0, b] cos(x) dx.
Integrating cos(x) with respect to x gives us the anti-derivative sin(x). Evaluating this anti-derivative at the upper limit b and lower limit 0, we have:
Area of S = [sin(x)] from 0 to b = sin(b) - sin(0) = sin(b).
Now, we set the areas of R and S equal to each other:
b^2/2 = sin(b).
To find the value of b that satisfies this equation, we can use numerical methods or graphing calculators to find the intersection point of the two functions f(b) = b^2/2 and g(b) = sin(b). The solution to this equation is approximately b = 1.404.
Therefore, the value of b for which the areas of regions R and S are equal is approximately 1.404.