# 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

Notice that $$y=\sqrt{1+x^3}$$ is defined only in $1+x^3\ge0\iff x\ge -1$.
So, if we divide the region into two small regions, then we can get an answer. $$\int_{-4}^{-1} \left(\frac 12x+2\right)dx+\int_{-1}^{2}\left(\frac 12x+2-\sqrt{1+x^3}\right)dx$$