I have the two equations:
y = 1 - 2x^2
y = abs(x)I solved the two equations using the absolute value for:
abs(x)and
-abs(x)Using these positive and negative absolute values I solved for the equations and found:
x =+- 1/2 x= +- 1First of all, are these x values correct and if so, am I supposed to have 2 separate integrals?
Between:
-1/2 < x < -1 and 1/2 < x < 1
Then do I add the integrals together? I am a bit confused as to the process that I am supposed to follow.
$\endgroup$1 Answer
$\begingroup$Hint: Clearly sketch the two equations in a single graph. You did find the points of intersection of these two graphs correctly. They do occur at $x = \pm 1/2$.
Now the easiest thing to do from here is to exploit symmetry. If you find the area between the curves on the interval $\lbrack 0, 1/2 \rbrack$, then it will be the same as the area on the interval $\lbrack -1/2, 0 \rbrack$. Hence, it is good enough to just find the bounded area on the interval $\lbrack 0, 1/2 \rbrack$ and multiply our result by $2$.
So how do we find the area bounded by those two curves bounded by $\lbrack 0, 1/2 \rbrack$? Well, first note that we can replace $y = abs(x)$ with simply $y = x$ since we are working on this interval. Now think subtracting two integrals.
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