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tandoorichicken
Jan5-04, 09:21 PM
I need to find the area between y=x and y=x^2
So this is what I did:
A = \int (x^2-x) \,dx
Then I found the limits of integration x=0 and x=1 because thats where the two graphs intersect
A = \int^1_0 (x^2-x) \,dx
I ended up with an answer of -1/6
What did I do wrong?

ShawnD
Jan5-04, 10:18 PM
intersect points:
x^2 = x

x^2 - x = 0

(x)(x-1) = 0

x = 1, x = 0



The upper limit is the line y = x, the lower limit is y = x^2

A = \int^1_0 x \,dx - \int^1_0 x^2 \,dx

A = \frac{x^2}{2} |^1_0 - \frac{x^3}{3} |^1_0

A = \frac{1^2}{2} - \frac{1^3}{3}

A = \frac{1}{2} - \frac{1}{3}

A = \frac{1}{6}



Your answer seems fine to me.

Hurkyl
Jan5-04, 10:20 PM
Which one's bigger?

tandoorichicken
Jan5-04, 10:53 PM
oh...... I get it