Find the area of the shaded region ONLY that lies between the line y = 1 y=1 & f ( x ) = sin 2 x f\left(x\right)=\sin2x .

this problem, we're asked to find the area of the shaded region only that lies between the line y equals one and the function f of x equals sine of two x. Now we can see this shaded area in yellow here on our graph. And we know that in order to find this area, we're gonna have to set up some definite integral where we take our top function and subtract that bottom function. Now because we're not given the bounds of this integral explicitly, we need to find them by setting our two functions equal to each other. So we wanna find these two intersection points of our function f of x and the line y equals one in order to bound our integral. So I'm gonna take the sine of two x and set it equal to one. So we need to think about our unit circle here. Where is the sine equal to one? Well, it's equal to one when whenever that argument when whatever that argument is, is equal to pi over two. Now because we need to think about multiple points where this happens, this is pi over two plus two pi n because we know that this is multiples of that pi over two that makes this argument equal to one. So we need to go ahead and divide by two on both sides here in order to get x fully by itself. So this becomes x equals pi over four and then plus pi n because we cancel that two out there. So these represent our intersection points. Now between these two functions, there really is an infinite number of intersection points, but because we're just looking at this shaded area, we only need these first two intersection points between zero and three pi over two. So I know that one of these is going to be pi over four. So that's that first time when the sine of two x is equal to one. We can see that based on this. If n is equal to zero, that makes this first point x equals pi over four. So when is the next time that sine of two x is equal to that one? Well, if we add a pi into that, so if x is equal to pi over four plus pi times one, just taking that first iteration there, that's going to give us five pi over four. So if that's our next intersection point on our graph, that represents this point here. And these two values are what's going to be the bounds of our integral. So we have x equals pi over four and x equals five pi over four. So now we can go ahead and set up that area integral. So here the area is gonna be equal to the definite integral from pi over four to five pi over four of our top function. In this case, y equals one. So just one minus our bottom function f of x, which is the sine of two x. So one minus the sine of two x d x, this integral represents our area. So now we just need to evaluate this integral in order to get our final numerical answer of that area. So starting off with this integral, the antiderivative of one is just going to be x. So this is x and then a minus this sign of two x. Technically, we would have to do a u substitution. You may be familiar with this process, but thinking about what we would have to do here if we set u equal to two x, that makes DU equal to two DX. So we would have to multiply that by two and one half, making this answer negative one half cosine of two X. Then this is bounded from Pi over four to five pi over four. So now that we have this antiderivative, I'm going to go ahead and put those two negatives together. This is x plus one half cosine of two x. Now we want to go ahead and apply those bounds using the fundamental theorem of calculus. So plugging in that first bound or that upper bound five PI over four, this is five PI over four plus one half times the cos of two times five Pi over four. Make sure you don't forget that two that's in our argument there. Then we are subtracting off that lower bound plugged in. So that's Pi over four plus one half times the cosine of two times Pi over four. So that's just Pi over two. So now that we're here, we want to go ahead and think about what these trig values are. If I take the cosine of 10 Pi over four, so I've multiplied that two, This is just going to be zero. So this whole term is zero leaving me with just that five pi over four. Then the same thing happens with this term two pi over four is really just pi over two. The cosine of pi over two is zero so that whole term goes to zero leaving me with just that pi over four. So if I take five pi over four and I subtract pi over four, that gives me four pi over four. Then those fours cancel leaving me with just pi. So this shaded area here is just equal to pi. If you have any questions here feel free to let us know in the comments and I'll see you in the next video.

9. Graphical Applications of Integrals

Area Between Curves

Calculus

9. Graphical Applications of Integrals

Area Between Curves

Difficoltà con Calculus?

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Find the area of the shaded region ONLY that lies between the line $y equals 1$ & $f of x equals sine 2 x$ .

2.43

1.73

$π$

$2 pi$

Guida verificata passo dopo passo

Identify the functions that bound the shaded region. Here, the upper function is y = 1 and the lower function is f(x) = sin(2x).

Determine the points of intersection between y = 1 and f(x) = sin(2x) to find the limits of integration. Solve the equation sin(2x) = 1 to find these points.

Set up the integral to find the area of the shaded region using the formula A = ∫[a, b] (f(x) - g(x)) dx, where f(x) is the upper function and g(x) is the lower function.

Substitute the functions into the integral: A = ∫[a, b] (1 - sin(2x)) dx, where a and b are the points of intersection found in step 2.

Evaluate the integral to find the area. This involves integrating each term separately and applying the limits of integration to find the definite integral.

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Find the area of the shaded region ONLY that lies between the line y=1y=1 & f(x)=sin2xf\(\left\)(x\(\right\))=\(\sin\)2x.

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Find the area of the shaded region ONLY that lies between the line $y equals 1$ & $f of x equals sine 2 x$ .