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When finding area bounded by 2 lines/curves by integration, I believe I 1st need to find the intersection point


$y=1-x$, $y=\sqrt{1+x}$

$1-x = \sqrt{1+x}$



So $x=0 \text{ or } -3$

Question is can I say

$$\text{area} = -\int^0_{-3}{(x^2+3x) \space dx}$$

I got a feeling its wrong? But why? It appears to me like I am "merging" 2 lines/curves, but that feels wrong ...

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up vote 1 down vote accepted

Let's observe picture bellow.Total area $A$ is represented as sum of blue and yellow area:

$A=A_1+A_2$ , where $A_1=\int\limits_{-1}^{0} \sqrt{1+x}\, dx$ ,and $A_2=\int\limits_{0}^{1} (1-x) \, dx$

So we have that:


enter image description here

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Your two curves only intersect once, at $(0,1)$. The "root" at $x=-3$ (which should be $3$) is extraneous, caused by squaring both sides. This means there is no region bounded by the two curves. If there were, the approach would be correct. For example, take $y=1-x$ and $y=1-x^2$. These intersect at $(0,1)$ and $(1,0)$ and the area between them would be $\int_0^1(x^2-x)\;dx$

Added: No, if you plot the two curves I gave there is a lens-shaped area between the two curves. If you want the area between two curves you do $\int y_2-y_1 \;dx$. The area between a curve and the x-axis is the special case $y_2=0$

In your original problem it does make sense to talk of the area bounded by the $x$ axis and the two curves. You could see this plot. $\sqrt{1+x}$ rises above the axis at $x=-1$. The two curves intersect (as you found) at $(0,1)$, then $1-x$ hits the axis at $x=1$. So your integral would be $$area=\int_{-1}^0 \sqrt{1+x} \;dx + \int_0^1(1-x)\;dx$$

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I was thinking along the lines of calculating the area bounded by the derived line and the x axis. Is that what you're doing in that example? Will that work for all equations? Or isit a "special case" – Jiew Meng Oct 24 '11 at 14:15

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