Find Area Between y=x and y=x^2 | -1/6 Answer

[tandoorichicken]

I need to find the area between y=x and y=x^2
So this is what I did:
[tex] A = \int (x^2-x) \,dx [/tex]
Then I found the limits of integration x=0 and x=1 because that's where the two graphs intersect
[tex] A = \int^1_0 (x^2-x) \,dx [/tex]
I ended up with an answer of -1/6
What did I do wrong?

 

[ShawnD]

intersect points:
[tex]x^2 = x[/tex]

[tex]x^2 - x = 0[/tex]

[tex](x)(x-1) = 0[/tex]

[tex]x = 1, x = 0[/tex]



The upper limit is the line [tex]y = x[/tex], the lower limit is [tex]y = x^2[/tex]

[tex]A = \int^1_0 x \,dx - \int^1_0 x^2 \,dx [/tex]

[tex]A = \frac{x^2}{2} |^1_0 - \frac{x^3}{3} |^1_0[/tex]

[tex]A = \frac{1^2}{2} - \frac{1^3}{3}[/tex]

[tex]A = \frac{1}{2} - \frac{1}{3}[/tex]

[tex]A = \frac{1}{6}[/tex]



Your answer seems fine to me.

 

[Hurkyl]

Which one's bigger?

 

[tandoorichicken]

oh... I get it