$f(x)=xe^{-\sqrt x}$

Find the value of $c$, such that the area bounded between the graph, the $x$-axis, $x=c$, and $x=c+1$ is maximized. Find the maximum area.

I don't know where to start with this one. I need help finding the maximum area.

  • $\begingroup$ Do you know how to integrate your function? If so, then integrate with bounds c to c+1. You will have a function of c. Derive it and find maximum. Good luck! $\endgroup$ – Galc127 Feb 12 '16 at 5:36
  • $\begingroup$ Yes I can find an integral. I suppose c, c+1 represent some number. That is why they are bounds. Thanks $\endgroup$ – User 210 Feb 12 '16 at 5:42

\begin{align*} F(x) &= \int f(x) \, dx \\ &=-2e^{-\sqrt{x}} \left( x^{3/2}+3x+6\sqrt{x}+6 \right) \\ A(c) &= \int_{c}^{c+1} f(x) \, dx \\ &= F(c+1)-F(c) \\ A'(c) &= f(c+1)-f(c) \\ \therefore \: \: 0 &= (c+1)e^{-\sqrt{c+1}}-ce^{-\sqrt{c}} \\ c & \approx 3.526 \\ A''(c) &= e^{-\sqrt{c+1}} \left( 1-\frac{\sqrt{c+1}}{2} \right)- e^{-\sqrt{c}} \left( 1-\frac{\sqrt{c}}{2} \right) \\ A''(3.526) & \approx -0.017 \\ & < 0 \end{align*}

The root gives maximal area.


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