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anonymous
 one year ago
Find the area of the region bounded by the functions f(x) = x^4 and g(x) = 2 − x^2.
anonymous
 one year ago
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
 one year ago
Best ResponseYou've already chosen the best response.0i've found the area bounded by the xaxis and f(x) its 3.771123

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would i have to integrate with respect to the y axis on this one?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1have you plotted this can you see the area enclosed by these curves?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep iv graphed it, i know what it looks like

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you are looking for the area bounded by the curves so you need to know where they meet to establish the limits of integration

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you could draw it using the Draw function here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay dont know how to draw function but the graphs seem to come together at x =1 and x=1

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so firstly integrate the functions within those limits and see what comes out

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.1It seems you need to use the method of slicing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which function f(x)or g(x)

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.1this looks like a disk

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is their a formula to this method ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1@blackstreet23 are you deliberately trying to take this of course? this is a simple integration.

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.1no i just want to help

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438814403525:dw

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.1ohh yeah i am sorry it is not revolving haha

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1intersections at (1'1) and (1,1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes okay i got that now what ....?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{?}^{?}( 2x^2)(x^4)\]

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.1if i am right this is because it is suppose the areas under the curve

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438814688554:dw "just" integrate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so for the boundris it would just be 1 to 1 right ?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.1if you see on the drawing that one of the users drew the upper function is 2x^2an

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.1and the lower function is X^4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay okay that gave me 44/15

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1yes, that is right limits are \(1 \le x \le +1\)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1integrate 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 yaxis multiply this result by 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why do i have to multiply by 2 ?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1because i integrate between 0 and 1 but its more direct to use limits 1 and 1

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.1but why should it be integrated separately if the bounded region is above the xaxis?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1he 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
 one year ago
Best ResponseYou've already chosen the best response.0okay the answer is 44/15 either way

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1yes  i dont know why i did it the longer way Irish Boy is right both methods will gave same answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay thank you very much :D

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\(\huge A = \int_{1}^{1} 2  x^2  x^4 \ dx\) \( \huge = [2x  \frac{x^3}{3}  \frac{x^5}{5} ]_{1}^{1}\) \(\huge = 2 [2x  \frac{x^3}{3}  \frac{x^5}{5} ]_{0}^{1}\) \(\huge = 2 [2  \frac{1}{3}  \frac{1}{5} ] = 2(\frac{30  5  3}{15}) = 2( \frac{22}{15}) = \frac{44}{15}\) \(\large \checkmark\)
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