Trigo Area Bounded.

Jacob

Jacob

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Jan 27, 2019
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Hi, I need help with 2 things. (I found the graph for this solution by using graphing calculator)

1. May I know how I can find the x-intercept = 5π/6? I found π/6 by letting sinx=0.5 and solve for x but Idk how I can find 5π/6.
2. Is my integral form correct to find the area of A?

My goal is to find the Area A+B+C.
1617630631945.png
 
Jacob said:
Hi, I need help with 2 things. (I found the graph for this solution by using graphing calculator)

1. May I know how I can find the x-intercept = 5π/6? I found π/6 by letting sinx=0.5 and solve for x but Idk how I can find 5π/6.
2. Is my integral form correct to find the area of A?

My goal is to find the Area A+B+C.
View attachment 26196
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The intersections (not really an x-intercept) are solutions of \(\sin(x)=\frac {1}{2}\). You know that one solution is \(\frac {\pi}{6}\), which is one of the special angles.

You should know that the two distinct solutions of such an equation are supplements of one another; this can be easily seen from the unit-circle representation of the sine, or from the symmetry of its graph about the line \(x = \pi\). And the supplement of \(\frac {\pi}{6}\) is \(\pi-\frac {\pi}{6}=\frac {5\pi}{6}\).

Your integrand should be the difference between the two functions, namely \(0.5 - \sin(x)\). Do you see why? It certainly can't be their product, as you wrote.
 
Dr.Peterson said:
Your integrand should be the difference between the two functions, namely \(0.5 - \sin(x)\). Do you see why? It certainly can't be their product, as you wrote.
Click to expand...
Is it because then when I integrate I will be getting positive value?

or Upper - Lower is the reason why 0.5 -sin (x)
 
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Jacob said:
Is it because then when I integrate I will be getting positive value?

or Upper - Lower is the reason why 0.5 -sin (x)
Click to expand...
Yes, both are true.

Ultimately, though, the reason is that to find an area, you integrate the element of area, which is height (upper minus lower in this case) time the width (dx in this case).
 
Therefore this is the solution for A area. Correct me if I'm wrong.

1617635662551.png
 
you have +cos(0) yet you replaced that with 1. Why is that?

1.12782 - 1 is exactly 0.12782, so why do you but the approximate symbol. Maybe you approximated before you put your approximate symbol but again 1.12782-1 = .12782 exactly. If you did approximate before, then put the approximate symbol there.
 
Jomo said:
you have +cos(0) yet you replaced that with 1. Why is that?

1.12782 - 1 is exactly 0.12782, so why do you but the approximate symbol. Maybe you approximated before you put your approximate symbol but again 1.12782-1 = .12782 exactly. If you did approximate before, then put the approximate symbol there.
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Isn't Cos(0) = 1?
 
Jomo said:
Yes, so why did you replace +cos(0) with -1? Maybe it should not have said +cos(0)?????
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Oh because,
[0.5(π/6)+cos(π/6)] - [0.5(0)+cos(0)]
[1.27282] - [1] = 0.12782

This is how I get 0.12782
 
Can someone tell what am I doing wrong when finding the area B? I'm getting negative as my final answer


1617638243398.png
 
Jacob said:
Can someone tell what am I doing wrong when finding the area B? I'm getting negative as my final answer


View attachment 26198
Click to expand...
The reason you broke the area into three parts was that which curve is on top changes from one to the next.

Always think! In this region, which is the top??? So what should the integrand be?
 
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