1
$\begingroup$

The region is enclosed by the curves $y=\sqrt{x+2}$, $y=\frac1{x+1}$, and lies between $x=0$ and $x=2$. Help please, any work would be helpful. Thank you!

$\endgroup$
  • 1
    $\begingroup$ Any work (of your own) would be helpful in understanding what you are stuck on. Have you drawn the region? $\endgroup$ – Matthew Leingang Dec 14 '14 at 17:05
  • 1
    $\begingroup$ Dang, I submitted my edit just after Sujaan's.... $\endgroup$ – teadawg1337 Dec 14 '14 at 17:06
1
$\begingroup$

Integrate the difference between 0 and 2:

f1 = Sqrt[x + 2];
f2 = 1/(x + 1);
Plot[{f1, f2}, {x, 0, 2}, Filling -> {1 -> {{2}, {None, LightGray}}}, 
       PlotTheme -> "Detailed", PlotLegends -> {f1, f2}]

Mathematica graphics

Integrate[f1 - f2, {x, 0, 2}]

Mathematica graphics

N[%]

Mathematica graphics

$\endgroup$
0
$\begingroup$

Hint. Since, for $x$ such that $0\leq x\leq2$, we have $\displaystyle \sqrt{x+2} > \frac1{x+1}$, then the desired area $A$ is such that $$ A =\int_0^2 \sqrt{x+2} \:dx-\int_0^2 \frac1{x+1} \:dx $$ The second integral on the right hand side is easy to find out, for the first integral just write $\displaystyle \sqrt{x+2}=(x+2)^{1/2}$.

$\endgroup$

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.