I've been having a few issues with this question and I don't know how to go about finding the enclosed area between the curve and the $x$-axis
$$y = x^2 + 2x - 3,\quad 0 \le x \le 2$$
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$\begingroup$ That would simply be the integral of the function, with respect to x, under the limits $x = 0$ and $x = 2$. $\endgroup$ – Gummy bears Sep 19 '15 at 4:53
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Here $y=(x^2+2x-3) = (x^2+2x+1)-4 = (x+1)^2-4$
Now drawing Graph of $y=(x+1)^2-4\Rightarrow (y+4) = (x+1)^2$
So it is an Upward parabola whose vertices's is at $(-1,-4)$
So our required area is $\displaystyle \int_{0}^{2}ydx\ = \left|\int_{0}^{1}ydx\right|+\int_{1}^{2}ydx$
Bcz in $x=0$ to $x=1,$ area bounded by curve and $X$ axis is below $X$ axis, Thats why we will take modulus sign
and in $x=01$ to $x=2,$ area bounded by curve and $X$ axis is above $X-$ axis
So for total area, We will add these two parts.
answered Sep 19 '15 at 4:59
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Hint: We have: $S = \displaystyle \int_{0}^1(0-(x^2+2x-3))dx + \displaystyle \int_{1}^2 (x^2+2x-3)dx$
