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I have 2 similar problems. The first one:

Find the area of the region bounded by the parabola $y = x^2$, the tangent line to this parabola at $(1,1)$ and the $x$-axis.

I set up my equation like this:

$$\int_0^1 2x - 1 - x^2 \, dx$$

Is that correct?

EDIT It is not correct. I need to use horizontal integration right? So it would be:

$$\int_0^1 \frac{y+1}{2} - \sqrt{y} \, dx$$

  1. Find the values of $c$ such that the area of the region bounded by the parabolas $y = x^2 - c^2$ and $y = c^2 - x^2$ is $576$.

I don't even know how to set this up. Any hints?

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  • $\begingroup$ 1. Integrate w.r.t. $y$: $$\int_0^1 \frac{y+1}{2} - \sqrt{y} \, dy$$ 2. Treat $c$ as a constant, draw a graph (symmetric about both axes), then set up integral. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 15 '18 at 18:25
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You're not going to get anywhere if your approach to problems is a hit-and-miss "can we try vertical integration here?" "okay, what about horizontal integration?"

Instead, you should actually understand what the region looks like, and then use that to set up an integral for the area.

In this case, the parabola and its tangent line look like

enter image description here

and this should tell you why your first integral with respect to $x$ doesn't work: it includes the entire area between the blue and orange curves, which also counts a large triangle below the $x$-axis.

So you have the following approaches:

  1. Split up the region. For $0 \le x \le \frac12$, we want the area between the parabola and the $x$-axis, which is $\int_0^{1/2} x^2\,dx$. For $\frac12 \le x \le 1$, we want the area between the parabola and the tangent line, which is $\int_{1/2}^{1} x^2-(2x-1)\,dx$. Then we add them together.
  2. Take the region between the two curves, then subtract the triangle. The bigger region has area $\int_0^1 x^2-(2x-1)\,dx$, and the triangle is a right triangle with sides $\frac12$ and $1$ , so its area is $\frac14$. (The triangle area could also be an integral if you wanted it to be.)
  3. Take the horizontal integral between $x = \sqrt y$ (blue curve) and $x= \frac{y+1}{2}$ (orange curve) for $0\le y \le 1$, which is $\int_0^1 \frac{y+1}{2}-\sqrt y \,dy$.
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