<p>I keep getting 5/12 for this problem, but the answer is 1/12. could anyone just set up the integral so i can check my work?</p>
<p>-find the area of the region bounded by y = x^2, the tangent line to the parabola at (1,1) and the x-axis.</p>
<p>I think that the trick is keeping in mind that you can't just take integral(x^2 - 2x +1)dx from x=0 to x=1 because you need the area between the two graphs above the x axis. Try integral(x^2)dx from x=0 to x=1/2 plus integral(x^2-2x+1)dx from x=1/2 to x=1. (1/2 is the value of x for which the linear part - the tangent line - equals zero).</p>
<p>sl8, I tried that and also the intedral with dy. did you get 1/12 for the answer?</p>
<p>y = x^2 call it f(x)</p>
<p>f'(1) = 2(1) = 2 = slope of tangent line</p>
<p>y - 1 = 2(x-1)
y = 2x - 1 <--this is your equation for the line lets call it g(x)</p>
<p>(Integral from 0 to 1 of f(x)dx ) - (Integral from .5 to 1 of g(x)dx) should be (1/3) - (1/4) = 1/12</p>
<p>Yeah, I got 1/12</p>
<p>[integral(x^2)dx from x=0 to x=1/2] = 1/24
[integral(x^2-2x+1)dx from x=1/2 to x=1] = 1/24</p>
<p>So for my final answer I just added them.</p>
<p>thanks ninos.</p>
<p>If you just take integral(x^2 - 2x +1)dx from x=0 to x=1, then does that also include the area between upper x-axis and 2x-1 from x=0 to x=1/2? It's a little confusing.</p>
<p>Yeah, I think that it does. Either that, or I just unnecessarily wasted time by breaking up the integral.</p>