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?

  • 1
    $\begingroup$ 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

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. $$\int_{-4}^{-1} \left(\frac 12x+2\right)dx+\int_{-1}^{2}\left(\frac 12x+2-\sqrt{1+x^3}\right)dx$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.