The region R is bounded by the x-axis and the graphs of y=2-x^3 and y=tan(x)

- anonymous

The region R is bounded by the x-axis and the graphs of y=2-x^3 and y=tan(x)

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- anonymous

I need to find the area of R

- anonymous

I got the graph part. I just need to find the area of that part.

- anonymous

this requires the coordinates of point P by solving the equation 2 - x^3 = tanx
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## More answers

- anonymous

how would you do that?

- anonymous

i figured you'd know...
I don't think there's an elementary algebra technique to solve that equation... you might need the help of a graphing calculator...

- anonymous

what part of the graphing calculator would help me with this?

- anonymous

finding the intersection of two graphs.... in this case y = tanx and y = 2-x^3

- anonymous

i would use zoom box on the graphing calculator

- anonymous

yes, but be care with using the functions of a graphing calculator... let the calculator find the intersection for you and not you zooming in on the intersection until you find it.
what kind of calculator are you using?

- anonymous

TI-84 PLus

- anonymous

How would I let calculator find the intersection points?

- anonymous

go to "y="
and in y1 enter tan (x)
y2 enter 2 - x^3

- anonymous

lemme know when you're done...

- anonymous

I did that then i hit zoom box until i got close enough then I used the trace button to get as close to the intersection

- anonymous

yeah that's what I mean about not doing that because the calculator can give you a better approximation of the intersection...

- anonymous

wait... make sure you're in "radian" mode.

- anonymous

yes im in radians and I entered the y1 and y2

- anonymous

ok, now press 2ND - CALC - 5 (intersect)

- anonymous

enter, enter, enter

- anonymous

o so thats how you do that!!!

- anonymous

calculator should give you
x = .90215
y = 1.2657
is that what you got?

- anonymous

yes so those points what go in my integral

- anonymous

yes. you'll use the lower limit of 0 and upper limit of 1.2657...

- anonymous

since you're using the calculator you might as well let the calculator find the area for you.. wanna see how it's done?

- anonymous

i know that part already but how you set up the problem for R bounded by the x-axis and S bounded by the y-axis

- anonymous

what's S? the problem only wanted region R as I put in the drawing....

- anonymous

tehn it asks for region S which is bounded by the y-axis at the same points

- anonymous

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that's region S? and you want the area for it?

- anonymous

yes

- anonymous

ok|dw:1333523054819:dw|
can we just take the whole area under y=2-x^3, take the area of S, then to get R, we subtract S from the whole area...
or do you want separate integrals for the area of R and S separately?

- anonymous

seperately because thats how my teacher wants it

- anonymous

ok, let's do R first then since that's what was asked first.
notice that the representative rectangle is laying down so the width of that rectangle is dy and the height is obtained by [right curve] - [left curve].

- anonymous

you must solve for x in both y=tanx and y=2-x^3....

- anonymous

so just put the integral in the calculator by (2-x^3)-(tan(x))dx

- anonymous

no, you must solve for x first.... in both curves....

- anonymous

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the reason you need to solve for x for both equations is because you need h to find the representative area dA.

- anonymous

so how you do that??
x=3squareroot of 2
tanx=?

- anonymous

|dw:1333523991360:dw|

- anonymous

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there's your area for R....
S should be easier because your taking dx and your height is top function - bottom function...

- anonymous

o i did it wrong

- anonymous

can you find area of dy by using the calculator

- anonymous

i got .729 un 2 so how you set up S

- anonymous

.729 for the area of R? lemme check...

- anonymous

yep that's what I got too... .729 :)

- anonymous

now S?

- anonymous

ok

- anonymous

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that's S on the lower left....