Find the area of the region bounded by the functions f(x) = x^4 and g(x) = 2 − x^2.

- anonymous

Find the area of the region bounded by the functions f(x) = x^4 and g(x) = 2 − x^2.

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- anonymous

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

- anonymous

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

- IrishBoy123

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

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## More answers

- anonymous

yep iv graphed it, i know what it looks like

- IrishBoy123

you are looking for the area bounded by the curves
so you need to know where they meet to establish the limits of integration

- IrishBoy123

you could draw it using the Draw function here

- anonymous

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

- IrishBoy123

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

- blackstreet23

It seems you need to use the method of slicing

- anonymous

which function f(x)or g(x)

- blackstreet23

this looks like a disk

- anonymous

is their a formula to this method ?

- IrishBoy123

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

- blackstreet23

no i just want to help

- welshfella

|dw:1438814403525:dw|

- blackstreet23

ohh yeah i am sorry it is not revolving haha

- welshfella

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

- anonymous

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

- blackstreet23

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

- blackstreet23

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

- IrishBoy123

|dw:1438814688554:dw|
"just" integrate

- anonymous

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

- blackstreet23

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

- blackstreet23

and the lower function is X^4

- anonymous

yay okay that gave me 44/15

- IrishBoy123

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

- welshfella

integrate 2 - x^2 between the limits 0 and 1
integrate x^4 between limits 0 and 1
subtract the second result from the first
then as the area is symmetrical about the y-axis multiply this result by 2

- anonymous

why do i have to multiply by 2 ?

- welshfella

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

- blackstreet23

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

- IrishBoy123

he changed the limits from -1, 1 to 0,1 as they are symmetrical, then x2 : but you can just do -1, 1 and fly with it

- anonymous

okay the answer is 44/15 either way

- welshfella

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

- anonymous

okay thank you very much :D

- welshfella

yw

- IrishBoy123

\(\huge A = \int_{-1}^{1} 2 - x^2 - x^4 \ dx\)
\( \huge = [2x - \frac{x^3}{3} - \frac{x^5}{5} ]_{-1}^{1}\)
\(\huge = 2 [2x - \frac{x^3}{3} - \frac{x^5}{5} ]_{0}^{1}\)
\(\huge = 2 [2 - \frac{1}{3} - \frac{1}{5} ] = 2(\frac{30 - 5 - 3}{15}) = 2( \frac{22}{15}) = \frac{44}{15}\)
\(\large \checkmark\)

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