Question

Area between the curves:
y=(x-2)^2 ; y=x

Answer 1

determine the intersection points if they are not defined

the integrate the difference of the functions across the interval

Answer 2

(1,1) & (4,4)

Answer 3

\[\int_{a}^{b}(x-2)^2-x~dx=\left.\frac13(x-2)^3-\frac12x^2\right|_{a}^{b}\]

Answer 4

So I need the area between 1 and 4?

Answer 5

yes

Answer 6

although I think amistre has the bottom and top curves flipped

Answer 7

i might; i tend to ignore the sign

Answer 8

5-2=3
2-5=-3

since the area is an absolute value; the sign is useless

Answer 9

\[A = -\frac{ 1 }{ 3 }x ^{3}+ \frac{ 5 }{ 2}x ^{2} - 4x | 4, 1\]

Answer 10

\[\frac13(4-2)^3-\frac124^2-(\frac13(1-2)^3-\frac121^2)\]
\[\frac83-8-(-\frac13-\frac12)\]
\[\frac83-8+\frac13+\frac12\]
is what i get

Answer 11

Mine is different than yours, I expanded the (x-2)^2 and than figured it all together. Was this wrong?

Answer 12

if you expanded correctly, then your way was fine

Answer 13

I must of did something wrong, your answer is correct except for the sign.

Answer 14

lol, story of me life ;)

Answer 15

haha

Answer 16

So I can either keep the equations the way they are and Intergrate each one OR I can combine them as long as I do it correctly?

Answer 17

(x−2)^2−x

x^2-4x-x+4

x^2-5x+4

\[\frac13x^3-\frac52x^2+4x\]

Answer 18

yes, the "doing it correctly" part is key ;)

Answer 19

Ain't that the truth.
Also, story of my life. I am understanding the Calculus for the most part but the Algebra is killing me at times.

Answer 20

\[\frac134^3-\frac524^2+4(4)-(\frac131^3-\frac521^2+4(1))\]
\[\frac{64}3-\frac{40}2+16-(\frac13-\frac52+4)\]
\[\frac{64}3-20+16-\frac13+\frac52-4\]
\[\frac{63}3-8+\frac52\]
with any luck thats right too

Answer 21

Thank you for your help.

Answer 22

good luck :) and yeah, it the algebra that bites