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March 29, 2015

March 29, 2015

Posted by **Melanie** on Friday, August 15, 2008 at 6:23am.

Since it says that the parabola passes through 1,1 can I assume that the line is y = x or is that completely wrong?

Doing that I got ∫ x - x^2 dx (where the interval from a to b goes from 0 to 1)

so the solved integral would be

x^2/2 - x^3/3] 0 to 1

= [(1)^2/2 - (1)3/3] - 0

= 1/2 - 1/3

= 1/6

What should I have done if it's wrong?

- Math integrals -
**Reiny**, Friday, August 15, 2008 at 9:47amThe tangent line is not y = x

Since the derivative of y = x^2 is dy/dx = 2x, at (1,1) dy/dx = 2, so the tangent line has a slope of 2

its equation would be y = 2x + b, with (1,1) on it

So, 1 = 2(1) + b, ------> b = -1

the tangent equation is y = 2x-1, which would result in an x-intercept of (1/2,0).

I would take the area between y = x^2 from 0 to 1 minus the right -angled triangle formed by the x-axis, y=2x-1 and x=1

let me know if you got 1/12.

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