Find the area of the region bounded by the curves y = x^2 − 1 and y = sin(x). Give your answer correct to 2 decimal places.

- anonymous

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- anonymous

not sure what to do about the graph being partially in the negative quadrants

- anonymous

i found the endpoints of intersection at -0.636733 and 1.40962 not sure where to go on from there

- anonymous

@ganeshie8 @dan815 @Michele_Laino @pooja195 @Kainui

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## More answers

- anonymous

do i have to integrate with respect to the y-axis ?

- IrishBoy123

hope this helps you

##### 1 Attachment

- anonymous

i had already graphed it :(

- IrishBoy123

cool
found where they intersect?

- anonymous

yep at x = -0.636733 and 1.40962

- anonymous

not sure what to do next ...

- IrishBoy123

well, next integrate :p
i'd be sure to keep the +ve and -ve signs in order so you might wish to do 2 integrations. they cross at x = 0. so integrate x = -0.636733 to 0, and x = 0 to 1.40962
all in dx
you can do dy if you want but you are creating a whole load of problems for yourself such as having to integrate arcsin

- anonymous

wouldn't the fact that some of the graph is on the negative quadrants give me a wrong answer

- IrishBoy123

yep, that's exactly what you have to look out for
i'll try draw something

- IrishBoy123

|dw:1438890387798:dw|

- IrishBoy123

it's a real drag but you will need a way to keep the sign consistent, the areas under the curve will come out as negative. i think you already know that. so you just have to deal with it.

- anonymous

well thats what i don't know how to do

- freckles

should be able to do integrate top-bottom

- freckles

y=sin(x) is always above y=x^2-1 on the given interval

- anonymous

wouldn't the integration include everything between the function and the red line too?
|dw:1438891297985:dw|

- freckles

so you want to include more area than shown?

- freckles

|dw:1438891485605:dw|
you want this area

- freckles

|dw:1438891497073:dw|not this

- anonymous

no, im just saying that if you find the integral of the top area it will find this area |dw:1438891538121:dw|
which includes some area that i don't want

- freckles

so you aren't convinced the area is given by:
the integration of the (top-bottom) on the given interval?
I may be able to help you understand that if that is what you don't understand

- anonymous

im just not clear on how you would find the area of the top precisely thats all

- freckles

|dw:1438891880460:dw|

- freckles

let me just start with this piece this is going to give you overlap

- freckles

you want to subract out that one part

- IrishBoy123

i am such a moron
shift the whole thing up by 1
integrating between \(y = x^2 -1\) and \(y=sin x\), "magnitude-wise", is same as integrating between \(y = x^2\) and \(y = sin x + 1\)
you will have same x intercepts too

- anonymous

you can do that !!!!!

- IrishBoy123

ask @freckles
sorry @freckles i had one of those "duh" moments......

- freckles

yeah and we should wind up with
\[\int\limits_{-0.637}^{1.141 }(\text{ top } - \text{ bottom }) dx \\ \int\limits_{-0.637}^{1.141 }(\sin(x) - (x^2-1)) dx \\ \]
but trying to think how to explain this in pieces...
\[\int\limits_0^{1.41 } \sin(x)dx-\int\limits_c^{1.41} (x^2-1) dx-\int\limits_0^c (x^2-1)-\int\limits_{-.637}^0 (x^2-1)dx--\int\limits_{-.637}^{0} \sin(x) dx \\ \text{ as we see this gives us } \\ \int\limits_{-.637}^{1.41} \sin(x) dx-\int\limits_{-.637}^{1.141}(x^2-1) dx \\ \text{ which can be written the way \above } \\ \int\limits_{-.637}^{1.41} (\sin(x)-(x^2-1) )dx\]
it is hard for me though to draw some of this overlapping stuff

- anonymous

okay so then it would be \[\int\limits_{-.636}^{1.409}\sin(x) +1 - x^2\] right?

- freckles

yes exactly you just do top - bottom

- anonymous

im getting -.376

- freckles

shouldn't be a negative number

- anonymous

for got the +1
1.6702

- anonymous

Thank you very much :D

- anonymous

one sec

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