Find the area of the shaded region between f ( x ) = sin 2 x \textcolor{orange}{f\left(x\right)=\sin2x} & g ( x ) = 2 sin x \textcolor{blue}{g\left(x\right)=2\sin x} from x = 0 x=0 to x = 2 π x=2\pi .

this problem, we're asked to find the area of the shaded region between f of x equals sine of two x and g of x equals two sine of x, specifically on the interval from x equals zero to x equals two pi. Now looking at our graph here, we can see that we're going to have to calculate two different integrals for these areas and add them together because our function two sine of x that starts out as the top function ends up on the bottom after these two functions cross paths. So let's go ahead and set up our integral here. So this is going to be the integral from a to b. So here we're looking at where that function g of x is on top. We can see that that goes from zero to Pi. So that first integral is from zero to Pi of that function g of x, which is two times the sine of x and then minus sine of two x. That's our function f of x. Then we are adding this together with the integral from pi to two pi. That's the other half of our integral here of net what is now our top function, the sine of two X minus two sine of X. So now that we have these integrals set up, we just need to evaluate them. So let's go ahead and do that here. Now the integral of this first term two sine of X is going to be a negative two cosine of X. And then for this next term here because we have that two on the inside here, this will end up giving us negative one half times the cosine of two X. And this is bounded from zero to pi. Now because we have the same exact terms happening here, I'm going to go through this quickly. This sine of two X is going to be negative one half times the cosine of two X and then minus a negative two cosine X, bounded from PI to two PI. I'm gonna go ahead and clean this up because I can make these double negatives positive. So this is negative two cosine of X plus one half cosine of two X, bounded from zero to Pi and then adding this together with negative one half cosine of two X plus two cosine of X bounded from Pi to two Pi. So now all that's left to do is actually plug in these bounds. So starting with our first interval here, we have a negative two times the cosine of pi plus one half the cosine of two times pi. So that's going to be two pi. So that's that first integral's upper bound plugged in. Now subtracting off our lower bound of that integral plugged in, this is going to be negative two cosine of zero plus a one half times the cosine of two times zero, which is still just zero. So all of this here is just that first integral with our bounds plugged in. Now we're adding this together with our second integral here. I'm going to go ahead and continue on this second line. So negative one half cosine of two X plugging in that top bound of two Pi. So this is negative one half times the cosine of two times two Pi. That is four Pi plus two times the cosine of two Pi. That is that upper bound of my second integral and then subtracting off here that lower bound plugged in negative one half cosine of two times Pi. So two Pi plus two times the cosine of Pi. So that is that lower bound. So here I have both integrals. So in that yellow, I have that first line of text there and then here with my second integral, that is this entire line. Now all of this stuff plugged in, we have to look at what all of these cosine values are. So what is the cosine of pi? It is negative one. The cosine of two pi is positive one. The cosine of zero is also positive one in both of these cases. And then the cosine of four Pi, that's just two rotations around my unit circle. That is also going to be positive one. Cosine of two Pi is still positive one. Cosine of two pi again positive one, and finally the cosine of pi that will be negative one. So replacing all of those cosine with their actual values here. In that first line we have negative two times negative one that is positive two, and then plus one half minus negative two times positive one, that is negative two, plus one half. So that's that first integral. And then adding this together, we're looking at this term now, negative one half times positive one is negative one half plus that two times one, which is just two, And then subtracting off your negative one half times one, that's still negative one half, plus two times negative one, that's negative two. So now once we combine all of these together, this value, these one halves are just going to cancel out and I'm just gonna be left with two plus two which is four. And then here the same thing will happen in that second interval. So this ends up being just four plus four which gives us a final answer of eight. And this eight represents the area between these two functions as shown on our graph here on the interval from zero to two pi. Now 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

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Multiple Choice

Find the area of the shaded region between $f of x equals sine 2 x$ & $g of x equals 2 sine x$ from $x equals 0$ to $x equals 2 pi$ .

Verified step by step guidance

Identify the points of intersection between the functions f(x) = sin(2x) and g(x) = 2sin(x) within the interval [0, 2π]. These points will be the limits of integration where the functions switch dominance.

Set the equations sin(2x) = 2sin(x) to find the points of intersection. Use trigonometric identities to simplify and solve for x.

Determine the intervals where f(x) is above g(x) and where g(x) is above f(x) using the points of intersection found in the previous step.

Apply the formula for the area between two curves: A = ∫[a to b] (f(x) - g(x)) dx + ∫[b to c] (g(x) - f(x)) dx, where a, b, and c are the points of intersection.

Evaluate the integrals separately over the determined intervals and sum the absolute values to find the total area of the shaded region.

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Struggling with Calculus?

Find the area of the shaded region between f(x)=sin2x\textcolor{orange}{f\left(x\right)=\sin2x} & g(x)=2sinx\textcolor{blue}{g\left(x\right)=2\sin x} from x=0x=0 to x=2πx=2\pi.

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Area Between Curves practice set

Find the area of the shaded region between $f of x equals sine 2 x$ & $g of x equals 2 sine x$ from $x equals 0$ to $x equals 2 pi$ .