Answer: 5.4142
Step-by-step explanation:
To find the area of the region enclosed by the two graphs, we can use an integral from x = 0 to [tex]x=\frac{3\pi}{4}[/tex] subtracting the bottom curve from the top curve.
First, we need to find which is the top curve and which is the bottom curve. See the attached image. This shows us that y = 4sin(x) is the top curve and y = 2cos(x) is the bottom curve.
Next, we can create our integral.
[tex]Area=\int\limits^b_a {(upper-lower)} \, dx[/tex]
[tex]Area=\int\limits^\frac{3\pi}{4} _0 {(4sin(x)-2cos(x))} \, dx[/tex]
Separate with the difference property:
[tex]Area=\int\limits^\frac{3\pi}{4} _0 {4sin(x)} \, dx - \int\limits^\frac{3\pi}{4} _0 {2cos(x)} \, dx[/tex]
Integrate with limits:
[tex]Area=-4cos(x)\vert^\frac{3\pi}{4} _0 - 2sin(x)^\frac{3\pi}{4} _0[/tex]
Apply the fundamental theorem of calculus:
[tex]Area=-4cos(\frac{3\pi}{4})+4cos(0)-( 2sin(\frac{3\pi}{4})-2sin(0))[/tex]
Find the values using the unit circle:
[tex]Area=-4(-\frac{\sqrt{2} }{2} )+4(1)-( 2(\frac{\sqrt{2} }{2} )-2(0))[/tex]
Distribute the -1:
[tex]Area=-4(-\frac{\sqrt{2} }{2} )+4(1)- 2(\frac{\sqrt{2} }{2} )+2(0)[/tex]
Multiply:
[tex]Area=2\sqrt{2} +4- \sqrt{2}[/tex]
Subtract:
[tex]Area=\sqrt{2} +4[/tex]
Compute:
[tex]Area\approx 5.4142[/tex]
You can also compute the integral with a calculator, if your teacher allows it.
