# Thread: I just want to ask if my answer on this problem (area of a plane region) is correct?

1. ## I just want to ask if my answer on this problem (area of a plane region) is correct?

Hi everyone. I tried solving this problem:

"Get the area of the region bounded by the upper branch of the parabola x=y2, the tangent line to the parabola at (1,1) and the x-axis."

Attached is a picture of my solution to the problem. I just want to ask if I did the right thing, and if not, what did I do wrong and how do I fix it?

Thank you very much for your help!

2. ## Re: I just want to ask if my answer on this problem (area of a plane region) is corre

You need to find the equation of the tangent line, and then find where it cuts the x-axis (say "a").

Then subtract the area under the upper branch of the parabola $\displaystyle y=\sqrt{x}$ (from 0 to 1) from the area under the tangent line from "a" to 1.

3. ## Re: I just want to ask if my answer on this problem (area of a plane region) is corre

Originally Posted by Debsta
You need to find the equation of the tangent line, and then find where it cuts the x-axis (say "a").

Then subtract the area under the upper branch of the parabola $\displaystyle y=\sqrt{x}$ (from 0 to 1) from the area under the tangent line from "a" to 1.
Thank you so much for your response! I managed to get the tangent line equation and "a". I'm wondering why I have to subtract the areas though? Based on what I can see on my notes, split areas area always added. Thank you in advance.

EDIT: Now I understand!

4. ## Re: I just want to ask if my answer on this problem (area of a plane region) is corre

kaonashi,

you don't have to split areas. You can subtract from right to left. Suppose you worked out the tangent line at (1, 1) to be $\displaystyle \ y = \dfrac{1}{2}x + \dfrac{1}{2}$.

Solve this for x in terms of y: $\displaystyle \ \$x = 2y - 1

Keep the form of $\displaystyle \ x = y^2 \$ corresponding to the parabola. From y = 0 to y = 1, the upper branch of the parabola $\displaystyle \ x = y^2 \$ is to the right of the tangent line $\displaystyle \ x = 2y - 1$.

The integrand is $\displaystyle \ [y^2 \ - \ (2y - 1)]dy. \$ The limits of integration are from y = 0 up to y = 1.

By factoring the quantity in the brackets, the antiderivative can be simplified to the point that when you substitute your limits of integration,
you will be doing the minimum of arithmetic.