Question

what is the area between y=x and y=(x^2/2)

Answer 1

so 4/3?

Answer 2

yes..

Answer 3

thats what i got using the integral process and it says its wrong

Answer 4

can you help me with similar problems

Answer 5

well, if u want using the integral process ...
first, find the interval x's between that curves
y=x^2/2
y=x
x^2/2 = x
x^2-2x=0
x(x-2)=0
for zeroes, satisfied for x=0 or x=2
now, calculate int(x-(x^2/2)dx from 0 to 2
= [1/2*x^2 - 1/6*x^3] from 0 to 2
= [1/2*(2)^2 - 1/6*(2)^3]-[1/2*(0)^2 - 1/6*(0)^3]
= (2-8/6) - 0
= 8/6 = 4/3

Answer 6

i know thats what i got using the integral process but my website for homework is marking it wrong

Answer 7

what about the area between x^2+6 and 12-x^2

Answer 8

wait still thinking was wrong with me for number one

Answer 9

yea, i got it...
2-8/6 = 12/6 - 8/6 = 4/6 = 2/3
it should be 2/3 not 4/3

Answer 10

thank you so much!

Answer 11

where did you get the 8/6 from? I got 2-4/6

Answer 12

1/6*(2)^3 = 1/6 * 8 = 8/6, right ?

Answer 13

oooooo i copied my notes down wrong

Answer 14

thx a lot! can you help with similar problems...i have a bunch im stuck on

Answer 15

for #2
do same idea like number one.
first, can u find the intersection of y=x^2+6 and y=12-x^2 (*just for x)
and then do integral process again :)