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area enclosed by y= root(1+x^3) , y=.5x+2, y=0
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Could someone set up the integral part?
well you'll need to find the points of intersection of the curves...
the upper graph is linear so its \[\int\limits_{0}^{a} (0.5x + 2) dx - \int\limits_{0}^{a} (\sqrt{1 + x^3}) dx\] where (a, f(a)) is the point of interesction of the 2 curves
I have the integral from -4 to 2, but when I plug it into the calculator it does not want to calculate.
so you need to use \[\int\limits_{-4}^{2}(0.5x + 2)dx - \int\limits_{-1}^{2}(\sqrt{1 + x^3}) dx\] because the domain of the 2nd function is restricted to x >= -1
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