Question

name: ____ date: ____ unit 8 practice packet lesson 8-6 1. write a set of integrals that represents the sum of all the areas of the shaded regions. use exact values for your boundaries, not rounded decimals. a. the figure shows the graph of ( y = 2cos(x) ), and the line ( y = sqrt{2} ), for ( -\frac{pi}{2} leq x leq \frac{pi}{2} ). b. the figure shows the graph of ( y = 2sin(x) ), and the line ( y = sqrt{3} ), for ( 0 leq x leq pi ). c. the figure shows the graphs of ( y = |x| ) and ( y = 3 + 2x - x^2 ) for ( -1 leq x leq 3 ). the ( x )-coordinates of the points of intersection of the graphs are ( x_1 ) and ( x_2 ) where ( x_1 < x_2 ). write a sum of integrals that represents the shaded regions. you do not need to solve for ( x_1 ) and ( x_2 ). 2. (calculator active) let ( r ) be the region bounded by the graph of ( y = e^{3x - x^3} ) and the horizontal lines ( y = 1 ) and ( y = 3 ), as shown in the figure. find the area of this bounded region.

Explanation:

Part a

Step 1: Find Intersection Points

To find where \( y = 2\cos(x) \) and \( y=\sqrt{2} \) intersect, solve \( 2\cos(x)=\sqrt{2}\Rightarrow\cos(x)=\frac{\sqrt{2}}{2} \). In the interval \( -\frac{\pi}{2}\leq x\leq\frac{\pi}{2} \), solutions are \( x = -\frac{\pi}{4} \) and \( x=\frac{\pi}{4} \).

Step 2: Determine Regions

• For \( -\frac{\pi}{2}\leq x<-\frac{\pi}{4} \), \( \sqrt{2}>2\cos(x) \), so area is \( \int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}}(\sqrt{2}-2\cos(x))dx \).
• For \( -\frac{\pi}{4}\leq x\leq\frac{\pi}{4} \), \( 2\cos(x)\geq\sqrt{2} \), so area is \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(2\cos(x)-\sqrt{2})dx \).
• For \( \frac{\pi}{4}

Step 1: Find Intersection Points

Solve \( 2\sin(x)=\sqrt{3}\Rightarrow\sin(x)=\frac{\sqrt{3}}{2} \). In \( 0\leq x\leq\pi \), solutions are \( x = \frac{\pi}{3} \) and \( x=\frac{2\pi}{3} \).

Step 2: Determine Regions

• For \( 0\leq x<\frac{\pi}{3} \), \( \sqrt{3}>2\sin(x) \), area: \( \int_{0}^{\frac{\pi}{3}}(\sqrt{3}-2\sin x)dx \).
• For \( \frac{\pi}{3}\leq x\leq\frac{2\pi}{3} \), \( 2\sin(x)\geq\sqrt{3} \), area: \( \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}}(2\sin x - \sqrt{3})dx \).
• For \( \frac{2\pi}{3}

Step 1: Analyze \( y = |x| \)

For \( -1\leq x<0 \), \( y=-x \); for \( 0\leq x\leq3 \), \( y = x \).

Step 2: Determine Regions

• For \( -1\leq x
• For \( x_1\leq x
• For \( x_2\leq x\leq3 \): \( |x| \) above \( 3 + 2x - x^2 \) (or vice - versa, from graph).

The sum of integrals is:
\( \int_{-1}^{x_1}((3 + 2x - x^2)-|x|)dx+\int_{x_1}^{x_2}((3 + 2x - x^2)-|x|)dx+\int_{x_2}^{3}(|x|-(3 + 2x - x^2))dx \)

Since \( |x|=-x \) for \( -1\leq x<0 \) and \( |x| = x \) for \( 0\leq x\leq3 \), we can also write it as:

For \( -1\leq x<0 \), \( |x|=-x \), so:

\( \int_{-1}^{x_1}((3 + 2x - x^2)-(-x))dx+\int_{x_1}^{x_2}((3 + 2x - x^2)-x)dx+\int_{x_2}^{3}(x-(3 + 2x - x^2))dx \)

Simplifying the integrands:

\( \int_{-1}^{x_1}(3 + 3x - x^2)dx+\int_{x_1}^{x_2}(3 + x - x^2)dx+\int_{x_2}^{3}(x^2 - x - 3)dx \)

Answer:

\( A=\int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}}(\sqrt{2}-2\cos x)dx+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(2\cos x - \sqrt{2})dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sqrt{2}-2\cos x)dx \)

Part b