1
$\begingroup$

Can someone help me with this? I have already found their intersections
$x = 1/2$ and $x = 1$.
I did this: $$\int_{\frac{1}{2}}^1 5x\ln(x) - (x+2)\ln(x) \; \mathrm d x$$ i conclude in:
$$ 2 \int_{\frac{1}{2}}^1 (x^2 - x)*\ln(x) - (x^2 -x)\; \mathrm d x $$ and i get $$ \frac{\ln({\frac{1}{2}}) - 1}{2}$$

but it's not the same that wolfram alpha shows here.

  • 0
    It would be much better if you type your solution here so we can spot the error if there's any – 2017-02-16
  • 0
    You can check that your answer is wrong without WolframAlpha, e.g. in an exam, because it is negative and an area should be positive. Review all the steps. – 2017-02-16
  • 0
    Hint: Integral $ln(x)$ and $x ln(x)$ can both be evaluated through the integration by parts method. – 2017-02-16
  • 0
    Could you show the antiderivative you obtained ? – 2017-02-16
  • 0
    Although it's shouldn't be that big of an issue (x+2) ln x is on top. – 2017-02-16

2 Answers 2

1

Well, it is seen from the graph that $(x+2)\ln x \geq 5x\ln x$ over $[0.5,1]$. Thus we have to find, $$\int_{0.5}^{1} (x+2)\ln x - 5\ln x \mathrm{d}x$$ $$= \int_{0.5}^{1} 2\ln x - 4x\ln x \mathrm{d}x$$ $$=\int_{0.5}^{1} -2(2x-1)\ln x \mathrm{d}x$$ $$= [-2(x^2-x)\ln x]|_{0.5}^{1} + 2\int_{0.5}^{1} (x-1) \mathrm{d}x$$ $$=\frac{1}{2}[\ln 2 -\frac{1}{2}] = \frac{1}{4}[\ln 4-1]$$

Hope it helps.

  • 0
    how did you get $ln(2)$ ? – 2017-02-16
  • 0
    @JuanIgnacioCuiule We have $\ln 0.5 = \ln 1 -\ln 2 = -\ln 2$. – 2017-02-16
  • 0
    Oh thanks now I get it – 2017-02-16
2

Well, the two intersections can be found by solving:

$$5x\ln\left(x\right)=\left(x+2\right)\ln\left(x\right)\space\Longleftrightarrow\space x=\frac{1}{2}\space\space\space\bigvee\space\space\space x=1\tag1$$

So, now in order to find the area between the two curves:

  • $$\mathcal{A}_1=\int_\frac{1}{2}^15x\ln\left(x\right)\space\text{d}x=\frac{5\left(\ln\left(4\right)-3\right)}{16}\tag2$$
  • $$\mathcal{A}_2=\int_\frac{1}{2}^1\left(x+2\right)\ln\left(x\right)\space\text{d}x=\frac{19}{16}-\frac{9\ln2}{8}\tag3$$

So, we get:

$$\mathcal{A}=\mathcal{A}_2-\mathcal{A}_1=\frac{19}{16}-\frac{9\ln2}{8}-\frac{5\left(\ln\left(4\right)-3\right)}{16}=\frac{\ln\left(4\right)-1}{4}\tag4$$