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Write the definite integral that computes the area of the region bounded by the graphs of $y=\sqrt{1+x^3}$, $y=\frac{1}{2}x+2$ and $y=0$.

I'm very confused because I graphed it on my calculator and it looked like it was infinite because the curve wasn't closed.. is it infinite or am I doing something completely wrong?

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    $\begingroup$ Looks to me like it should be x=0; certainly the square root is not defined for x < -1. $\endgroup$ – Keith Jan 3 '14 at 3:38
  • $\begingroup$ yeah I agree with you. I'm going to ask my teacher if that was a typo $\endgroup$ – sloth1111 Jan 3 '14 at 3:51
  • $\begingroup$ @sloth1111: see my answer. I think we can solve your question. $\endgroup$ – mathlove Jan 3 '14 at 3:56
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Notice that $$y=\sqrt{1+x^3}$$ is defined only in $1+x^3\ge0\iff x\ge -1$.

See the graph.

So, if we divide the region into two small regions, then we can get an answer. $$\int_{-4}^{-1} \left(\frac 12x+2\right)dx+\int_{-1}^{2}\left(\frac 12x+2-\sqrt{1+x^3}\right)dx$$

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  • $\begingroup$ ohhhhhhhh!! i didn't even think of dividing the area into 2 distinct pieces. that makes a lot of sense looking at the restrictions with the radical! thank you! $\endgroup$ – sloth1111 Jan 3 '14 at 4:08
  • $\begingroup$ You are welcome! $\endgroup$ – mathlove Jan 3 '14 at 4:42
  • $\begingroup$ @mathlove. Fortunately, they did not ask for the value. The antiderivative of Sqrt[1+x^3] is just awful to me. Do you know any change of variables which would make it nice ? $\endgroup$ – Claude Leibovici Jan 3 '14 at 6:25
  • $\begingroup$ @ClaudeLeibovici: A good point! I have no idea. wolframalpha.com/input/… $\endgroup$ – mathlove Jan 3 '14 at 6:30
  • $\begingroup$ @mathlove. Just for fun, ask WA just the antiderivative ! $\endgroup$ – Claude Leibovici Jan 3 '14 at 6:37

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