anonymous
  • anonymous
Find the area of the region bounded by the functions f(x) = x^4 and g(x) = 2 − x^2.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i've found the area bounded by the x-axis and f(x) its 3.771123
anonymous
  • anonymous
would i have to integrate with respect to the y -axis on this one?
IrishBoy123
  • IrishBoy123
have you plotted this can you see the area enclosed by these curves?

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More answers

anonymous
  • anonymous
yep iv graphed it, i know what it looks like
IrishBoy123
  • 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
  • IrishBoy123
you could draw it using the Draw function here
anonymous
  • anonymous
okay dont know how to draw function but the graphs seem to come together at x =1 and x=-1
IrishBoy123
  • IrishBoy123
so firstly integrate the functions within those limits and see what comes out
blackstreet23
  • blackstreet23
It seems you need to use the method of slicing
anonymous
  • anonymous
which function f(x)or g(x)
blackstreet23
  • blackstreet23
this looks like a disk
anonymous
  • anonymous
is their a formula to this method ?
IrishBoy123
  • IrishBoy123
@blackstreet23 are you deliberately trying to take this of course? this is a simple integration.
blackstreet23
  • blackstreet23
no i just want to help
welshfella
  • welshfella
|dw:1438814403525:dw|
blackstreet23
  • blackstreet23
ohh yeah i am sorry it is not revolving haha
welshfella
  • welshfella
intersections at (-1'1) and (1,1)
anonymous
  • anonymous
yes okay i got that now what ....?
blackstreet23
  • blackstreet23
\[\int\limits_{?}^{?}( 2-x^2)-(x^4)\]
blackstreet23
  • blackstreet23
if i am right this is because it is suppose the areas under the curve
IrishBoy123
  • IrishBoy123
|dw:1438814688554:dw| "just" integrate
anonymous
  • anonymous
okay so for the boundris it would just be -1 to 1 right ?
blackstreet23
  • blackstreet23
if you see on the drawing that one of the users drew the upper function is 2-x^2an
blackstreet23
  • blackstreet23
and the lower function is X^4
anonymous
  • anonymous
yay okay that gave me 44/15
IrishBoy123
  • IrishBoy123
yes, that is right limits are \(-1 \le x \le +1\)
welshfella
  • 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
  • anonymous
why do i have to multiply by 2 ?
welshfella
  • welshfella
because i integrate between 0 and 1 but its more direct to use limits -1 and 1
blackstreet23
  • blackstreet23
but why should it be integrated separately if the bounded region is above the x-axis?
IrishBoy123
  • 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
  • anonymous
okay the answer is 44/15 either way
welshfella
  • welshfella
yes - i dont know why i did it the longer way Irish Boy is right both methods will gave same answer
anonymous
  • anonymous
okay thank you very much :D
welshfella
  • welshfella
yw
IrishBoy123
  • 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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