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Sketch the region enclosed by the given curves and find its area.

$ y = \tan x $ , $ y = 2 \sin x $ , $ \frac{-\pi}{3} \le x \le \frac{\pi}{3} $

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$A=2-2 \ln 2$

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

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Lectures

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Sketch the region bounded …

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Sketch the region enclosed…

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Find the area of the regio…

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Let'S try to sketch the region close by those 2 curves and since the domain specified only need to jaw the curve inside the domain, so our x here is our x y plane. This is not good, so i said this is our x y plane, where x goes from negative pi over 3 to pi over 3 pi, so x equals poor 3 tangent x, equals to square rod. 3 x equal to negative power. 3 tangent x equals to negative square root. All right x, equals 0 is 0 and we know tangent axis at our function, so it looks like this and for, on the other hand, for the second curve or 2 sine x y know sine x looks like this. So we only need to find the end points, so endpoints x equals to pi over 3. We know 6 is a square 3 over 2 and we multiply by 2, so it will be square 3 and as we can see, those 2 curves intersect and those 2 end point a region because 2 sine x is also an odd function. So it looks like this all right and here it's already closed the region and as we mention everything here is symmetric by this region. So we know those 2 parts has have the exactly same area, so we only need to calculate 1 of them that multiplied by 2. So say i want to calculate this area. It will be integral with respect to x, since those 2 curves can be represented by x and of the boundary here is goes from 0 to 3 pi over 3 and the things that the integral will be the upper curve minus the lower curve, which is 2 Sine x, minus tangent, x, okay and antidotive of this is just minus 2 cosine x minus that are antidotangent. We know that anti dutie with tenant tangent x is negative. Log absolutly cos x, so this will be plus log cos x and vater ate at pi over 3 to 0, so x equals power 3. This is her. Half is just 2 times minus 1 plus log half minus x equal to 0. This is 2 times negative. 2 here because cosine 0 is 1 and we know log y, so minus 2 plus 0. This will return here is plus 4. Minus 2 is 2 and plus 2 times log 1 over 2. All right and for this log we can bring 1 negative sign to make this log 2 point. What i'm seeing here we have. This for a general log is like negative log: a over b equals to log b over a all right. It'S like that! You bring in the negative side to take the recipocal inside, so this also equals to 2 minus 2 log 2, both 2. Our answer,

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