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I need to find the area of R
I got the graph part. I just need to find the area of that part.
this requires the coordinates of point P by solving the equation 2 - x^3 = tanx |dw:1333521423411:dw|
how would you do that?
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...
what part of the graphing calculator would help me with this?
finding the intersection of two graphs.... in this case y = tanx and y = 2-x^3
i would use zoom box on the graphing calculator
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?
How would I let calculator find the intersection points?
go to "y=" and in y1 enter tan (x) y2 enter 2 - x^3
lemme know when you're done...
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
yeah that's what I mean about not doing that because the calculator can give you a better approximation of the intersection...
wait... make sure you're in "radian" mode.
yes im in radians and I entered the y1 and y2
ok, now press 2ND - CALC - 5 (intersect)
enter, enter, enter
o so thats how you do that!!!
calculator should give you x = .90215 y = 1.2657 is that what you got?
yes so those points what go in my integral
yes. you'll use the lower limit of 0 and upper limit of 1.2657...
since you're using the calculator you might as well let the calculator find the area for you.. wanna see how it's done?
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
what's S? the problem only wanted region R as I put in the drawing....
tehn it asks for region S which is bounded by the y-axis at the same points
|dw:1333522896129:dw| that's region S? and you want the area for it?
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?
seperately because thats how my teacher wants it
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].
you must solve for x in both y=tanx and y=2-x^3....
so just put the integral in the calculator by (2-x^3)-(tan(x))dx
no, you must solve for x first.... in both curves....
|dw:1333523725608:dw| the reason you need to solve for x for both equations is because you need h to find the representative area dA.
so how you do that?? x=3squareroot of 2 tanx=?
|dw:1333524156076:dw| there's your area for R.... S should be easier because your taking dx and your height is top function - bottom function...
o i did it wrong
can you find area of dy by using the calculator
i got .729 un 2 so how you set up S
.729 for the area of R? lemme check...
yep that's what I got too... .729 :)
|dw:1333524733187:dw| that's S on the lower left....