Shade the region bounded by f(x)=sinx\textcolor{blue}{f\left(x\right)=\sin x} & g(x)=cos2x\textcolor{orange}{g\left(x\right)=\cos2x} on the interval [π6,5π6]\left\lbrack\frac{\pi}{6},\frac{5\pi}{6}\right\rbrack. Then set up an integral to represent the region's area.

A = ∫ π 6 5 π 6 ( sin ⁡ x − cos ⁡ 2 x ) d x A=\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\left(\sin x-\cos2x\right)dx

Shade the region bounded by f ( x ) = sin x \textcolor{blue}{f\left(x\right)=\sin x} & g ( x ) = cos 2 x \textcolor{orange}{g\left(x\right)=\cos2x} on the interval [ π 6 , 5 π 6 ] \left\lbrack\frac{\pi}{6},\frac{5\pi}{6}\right\rbrack . Then set up an integral to represent the region's area.

this problem, we're asked to shade the region bounded by the two functions f of x equals sine of x and g of x equals cosine of two x on the interval from pi over six to five pi over six. Then we want to set up an integral in order to represent this region's area. So let's come over to our graph here. We can see that we have our function f of x, which is equal to the sine of x in blue here, and then we have g of x equals the cosine of two x. Now looking at this, we can see that where our function is on top versus on bottom kind of changes throughout this graph because they're going up and down as trig functions often do. But we're specifically looking at the interval from pi over six to five pi over six. So looking at my graph here, I can see that from pi over six to five pi over six my function f of x is on the top and my function g of x is on the bottom. So I can see that the area in between my two curves here is right in this middle region with f of x on top and g of x on bottom. So going ahead and shading this region with my yellow highlighter here. Now that we have shaded that region, we know what area we're looking for. We want to go ahead and set up an integral in order to find this area. Remember that in order to do this, we want a definite integral of that top function minus the bottom function. So setting up this integral area is equal to the integral from PI over six to five PI over six that was given to us in our problem here of that top function here is the sine of X minus our bottom function which is the cosine of two x integrated with respect to x. Now in this problem, we are specifically just asked to set up this integral. So this is our final answer here. If we wanted to actually find this shaded area, we could go ahead and evaluate this integral as well. Feel free to let us know if you have any questions and I'll see you in the next one.

Shade the region bounded by f ( x ) = sin ⁡ x \textcolor{blue}{f\left(x\right)=\sin x} & g ( x ) = cos ⁡ 2 x \textcolor{orange}{g\left(x\right)=\cos2x} on the interval [ π 6 , 5 π 6 ] \left\lbrack\frac{\pi}{6},\frac{5\pi}{6}\right\rbrack . Then set up an integral to represent the region's area.

A = ∫ π 6 5 π 6 ( sin ⁡ x − cos ⁡ 2 x ) d x A=\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\left(\sin x-\cos2x\right)dx

A = ∫ π 6 5 π 6 ( cos ⁡ 2 x − sin ⁡ x ) d x A=\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\left(\cos2x-\sin x\right)dx

A = ∫ π 6 5 π 6 ( sin ⁡ x − cos ⁡ 2 x ) d x A=\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\left(\sin x-\cos2x\right)dx

9. Graphical Applications of Integrals

Area Between Curves

Calculus

9. Graphical Applications of Integrals

Area Between Curves

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Multiple Choice

Shade the region bounded by $f (x) = \sin x$ & $g of x equals cosine 2 x$ on the interval $[\frac{π}{6}, \frac{5 π}{6}]$ . Then set up an integral to represent the region's area.

Graph showing the area between the curves f(x)=sin x and g(x)=cos 2x, shaded between π/6 and 5π/6.

$A = \int_{\frac{π}{6}}^{\frac{5 π}{6}} (\cos 2 x - \sin x) d x$

Graph showing the area between the curves f(x)=sin x and g(x)=cos 2x, shaded between π/6 and 5π/6.

$A = \int_{\frac{π}{6}}^{\frac{5 π}{6}} (\cos 2 x - \sin x) d x$

Graph showing the area between the curves of f(x)=sin x and g(x)=cos 2x, shaded between π/6 and 5π/6.

$A = \int_{\frac{π}{6}}^{\frac{5 π}{6}} (\sin x - \cos 2 x) d x$

Graph showing the area between the curves f(x)=sin x and g(x)=cos 2x, shaded between π/6 and 5π/6.

$A = \int_{\frac{π}{6}}^{\frac{5 π}{6}} (\sin x - \cos 2 x) d x$

Verified step by step guidance

Step 1: Identify the functions and the interval. The problem involves two functions: f(x) = sin(x) (blue curve) and g(x) = cos(2x) (orange curve). The interval of interest is [π/6, 5π/6].

Step 2: Determine which function is above the other within the interval. From the graph, f(x) = sin(x) is above g(x) = cos(2x) in the shaded region between x = π/6 and x = 5π/6.

Step 3: Set up the integral for the area. The area of the region is given by the integral of the difference between the upper function and the lower function over the interval. This is expressed as A = ∫[π/6, 5π/6] (sin(x) - cos(2x)) dx.

Step 4: Break down the integral. To compute the area, you would integrate sin(x) and cos(2x) separately over the interval [π/6, 5π/6]. Recall that the integral of sin(x) is -cos(x), and the integral of cos(2x) involves a substitution to account for the factor of 2.

Step 5: Evaluate the definite integral. After integrating, substitute the limits of integration (π/6 and 5π/6) into the resulting expressions to find the area. Simplify the result to obtain the final value of the area.

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