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

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

Mathematics
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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?

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Other answers:

yep iv graphed it, i know what it looks like
you are looking for the area bounded by the curves so you need to know where they meet to establish the limits of integration
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 \(-1 \le x \le +1\)
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
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?
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
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
\(\huge A = \int_{-1}^{1} 2 - x^2 - x^4 \ dx\) \( \huge = [2x - \frac{x^3}{3} - \frac{x^5}{5} ]_{-1}^{1}\) \(\huge = 2 [2x - \frac{x^3}{3} - \frac{x^5}{5} ]_{0}^{1}\) \(\huge = 2 [2 - \frac{1}{3} - \frac{1}{5} ] = 2(\frac{30 - 5 - 3}{15}) = 2( \frac{22}{15}) = \frac{44}{15}\) \(\large \checkmark\)

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