# definite integral area help

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?

• Looks to me like it should be x=0; certainly the square root is not defined for x < -1. – Keith Jan 3 '14 at 3:38
• yeah I agree with you. I'm going to ask my teacher if that was a typo – sloth1111 Jan 3 '14 at 3:51
• @sloth1111: see my answer. I think we can solve your question. – mathlove Jan 3 '14 at 3:56

## 1 Answer

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

• 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! – sloth1111 Jan 3 '14 at 4:08
• You are welcome! – mathlove Jan 3 '14 at 4:42
• @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 ? – Claude Leibovici Jan 3 '14 at 6:25
• @ClaudeLeibovici: A good point! I have no idea. wolframalpha.com/input/… – mathlove Jan 3 '14 at 6:30
• @mathlove. Just for fun, ask WA just the antiderivative ! – Claude Leibovici Jan 3 '14 at 6:37