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i've found the area bounded by the x-axis and f(x) its 3.771123

would i have to integrate with respect to the y -axis on this one?

have you plotted this
can you see the area enclosed by these curves?

yep iv graphed it, i know what it looks like

you could draw it using the Draw function here

okay dont know how to draw function but the graphs seem to come together at x =1 and x=-1

so firstly integrate the functions within those limits
and see what comes out

It seems you need to use the method of slicing

which function f(x)or g(x)

this looks like a disk

is their a formula to this method ?

@blackstreet23
are you deliberately trying to take this of course?
this is a simple integration.

no i just want to help

|dw:1438814403525:dw|

ohh yeah i am sorry it is not revolving haha

intersections at (-1'1) and (1,1)

yes okay i got that now what ....?

\[\int\limits_{?}^{?}( 2-x^2)-(x^4)\]

if i am right this is because it is suppose the areas under the curve

|dw:1438814688554:dw|
"just" integrate

okay so for the boundris it would just be -1 to 1 right ?

if you see on the drawing that one of the users drew the upper function is 2-x^2an

and the lower function is X^4

yay okay that gave me 44/15

yes, that is right
limits are \(-1 \le x \le +1\)

why do i have to multiply by 2 ?

because i integrate between 0 and 1
but its more direct to use limits -1 and 1

but why should it be integrated separately if the bounded region is above the x-axis?

okay the answer is 44/15 either way

yes - i dont know why i did it the longer way
Irish Boy is right
both methods will gave same answer

okay thank you very much :D

yw