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Mathematics 9 Online
OpenStudy (nathanjhw):

Consider the function. Use the notion of signed area to evaluate the following definite integral (-4,4) f(x) dx.

OpenStudy (nathanjhw):

OpenStudy (nathanjhw):

OpenStudy (nathanjhw):

2 – 2 pi 4 – pi 2 – 4 pi 4 – 4 pi 4 – 2 pi

OpenStudy (nathanjhw):

@Zarkon

OpenStudy (nathanjhw):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

how far did you get into this?

OpenStudy (nathanjhw):

I'm not sure where to start.

jimthompson5910 (jim_thompson5910):

why not break it up into 3 pieces and find the area of each piece

jimthompson5910 (jim_thompson5910):

like so

OpenStudy (nathanjhw):

-x-2=-16 -sqrt(4-x^2) = -6.283 - 8.588i x-2 = -16

jimthompson5910 (jim_thompson5910):

idk how you're getting that

OpenStudy (nathanjhw):

Im taking the integral of (-4,4) and filling in those values for f(x).

jimthompson5910 (jim_thompson5910):

i see

jimthompson5910 (jim_thompson5910):

break it up into 3 pieces

jimthompson5910 (jim_thompson5910):

first piece \[\Large \int_{-4}^{-2}(-x-2)dx\]

jimthompson5910 (jim_thompson5910):

second piece \[\Large \int_{-2}^{0}-\sqrt{4-x^2}dx\]

OpenStudy (nathanjhw):

-10

jimthompson5910 (jim_thompson5910):

third piece \[\Large \int_{2}^{4}(x-2)dx\]

jimthompson5910 (jim_thompson5910):

evaluate each integral

OpenStudy (nathanjhw):

-10 -pi 2

jimthompson5910 (jim_thompson5910):

exactly one of those that you listed is incorrect (the other two are correct)

OpenStudy (nathanjhw):

The first one is -2. Forgot the -x.

jimthompson5910 (jim_thompson5910):

hint: areas above the x axis are positive areas areas below the x axis are negative areas you can simply look and determine the sign of the area without having to do any math at all

OpenStudy (nathanjhw):

So is the answer: 2-2pi?

OpenStudy (nathanjhw):

Or am I jumping to conclusions?

jimthompson5910 (jim_thompson5910):

list the areas of each piece again

OpenStudy (nathanjhw):

-2 -pi 2

jimthompson5910 (jim_thompson5910):

the -2 is incorrect why? because the red triangle is ABOVE the x axis, so it has to be positive at least. It's some positive number. Try again

OpenStudy (nathanjhw):

2 -pi 2

jimthompson5910 (jim_thompson5910):

yep, now we add: (Area1)+(Area2)+(Area3) = (2) + (-pi) + (2) = 4 - pi

OpenStudy (nathanjhw):

Thanks. I was just making small mistakes, but I think I understand it now.

OpenStudy (nathanjhw):

Are you able to help me through another one?

jimthompson5910 (jim_thompson5910):

sure, just one more though

OpenStudy (nathanjhw):

Alright thanks! I think I mainly need help breaking it into pieces.

OpenStudy (nathanjhw):

The following graph of f(x) consists of line segments and semicircles. Use the graph and the notion of signed area to evaluate the definite integral (1,9) f(x) dx.

OpenStudy (nathanjhw):

OpenStudy (nathanjhw):

18 – 4pi 16 – 2.5pi 16 – 2pi 14 – pi 14 – 4pi

jimthompson5910 (jim_thompson5910):

break it up like this

jimthompson5910 (jim_thompson5910):

actually, you don't need one line in there, here's a more simpler version

OpenStudy (nathanjhw):

Now how do I find the integrals of all those pieces?

jimthompson5910 (jim_thompson5910):

well you can use plain old geometry instead of calculus to find the area of each piece

OpenStudy (nathanjhw):

So I think the first piece's area is pi/2= 1.57

OpenStudy (nathanjhw):

I used the area of a semicircle which is (pi x r^2)/2.(pi x 1^2)/2 =pi/2

OpenStudy (nathanjhw):

I then broke the second piece into two and made a square and a triangle.

OpenStudy (nathanjhw):

The area of the square is 4.

OpenStudy (nathanjhw):

The area of the triangle is (4 x 2)/2=4

OpenStudy (nathanjhw):

So the area of the whole second piece is 8.

OpenStudy (nathanjhw):

The area of the third piece is (6x2)/2 = 6

OpenStudy (nathanjhw):

The area of the last piece is (pi x 3^2)/2 = (pi x 9)/2 = 14.13

OpenStudy (nathanjhw):

All together the area of all the pieces is 1.57 + 8 + 6 + 14.13 = 29.7

jimthompson5910 (jim_thompson5910):

one sec

jimthompson5910 (jim_thompson5910):

you forgot about the 2x2 square that is below the first semicircle

OpenStudy (nathanjhw):

So then it's 33.7?

OpenStudy (nathanjhw):

Wait I need to minus the last semicircle.

OpenStudy (nathanjhw):

So i get 5.44. 1.57 + 4 + 8 +6+ + (-14.13).

OpenStudy (nathanjhw):

I don't think that is one of my answers @jim_thompson5910

OpenStudy (nathanjhw):

Did I do something wrong?

jimthompson5910 (jim_thompson5910):

let me see

jimthompson5910 (jim_thompson5910):

ok, two things 1) keep everything in terms of pi (don't approximate pi) 2) the semicircle under the curve has a negative area

OpenStudy (nathanjhw):

So the first semicircle is pi/2 and the second semicircle is -9pi/2. I think everything else should be the same.

OpenStudy (nathanjhw):

Should I multiple all the values by 2?

jimthompson5910 (jim_thompson5910):

why multiply by 2?

OpenStudy (nathanjhw):

So I get pi and -9pi.

OpenStudy (nathanjhw):

I'd get 36 -8pi = 4-2pi.

OpenStudy (nathanjhw):

or 18 - 4pi. Which is one of my answers.

OpenStudy (nathanjhw):

Yeah I think that's right.

OpenStudy (nathanjhw):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

18-4pi is correct

OpenStudy (nathanjhw):

Alright I have one more. I'm almost possible I know the answer, but do you mind if I just double check with you?

OpenStudy (nathanjhw):

It's pretty basic.

OpenStudy (nathanjhw):

The graph of f(x) is shown below. Which of the following are true?

OpenStudy (nathanjhw):

I only. II only. III only. II and III only. I and II only. <---- My answer

jimthompson5910 (jim_thompson5910):

one sec

jimthompson5910 (jim_thompson5910):

I agree with what you got. Nice job

OpenStudy (nathanjhw):

Thank you very much! I really appreciate you spending all this time helping me. I'd give you more medals if I could, but I can't. At the very least you have a fan.

jimthompson5910 (jim_thompson5910):

you're welcome

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