# Area between the curve and x-axis

Find the area between the curve $y={2\over x-1}-1$ and the $x$-axis over the interval $[2,4]$

Would this solve the question $\int_{2}^4( {2\over x-1}-1)dx$ ?

So...

$[2\ln(x-1)-x+c]_2^4$ would be incorrect because the curve would be below the x-axis for $3$ to $4$

• But be cautious, the function is negative on $(3, 4]$. – Zhanxiong Aug 11 '15 at 14:06
• Hi stuart, I see you have already asked a few questions, most of which have received good answers. If you're satisfied with an answer, you can not only upvote it, but also accept it by using the little check mark right below the up- and downvote arrows. This lets whoever answered your question know that you are indeed satisfied with the answer, and it helps people spot questions which still need more attention. :) – Huy Aug 11 '15 at 14:11

$$\int_{2}^4 \bigg({2\over x-1}-1\bigg)dx$$

Yes it will, you are right

$$=\underbrace{\int_{x=2}^{x=3}\int_{y=0}^{y=\frac{2}{x-1}}\,dy\,dx}_{=\ln(4)}-\underbrace{\int_{x=3}^{x=4}\int_{y=\frac{2}{x-1}}^{y=0}\,dy\,dx}_{=-\ln\big(\frac 9 4\big)}$$

$$\boxed{\color{blue}{=\ln(4)+\ln\big(\frac 9 4\big)\approx0.575}}$$

• perhaps your limits $1\to 3$ in the first part are wrong you may check if you don't mind – Harish Chandra Rajpoot Aug 11 '15 at 14:16
• it's correct, I will put more info – 3SAT Aug 11 '15 at 14:18
• Direct integration on the whole interval is not correct. – Zhanxiong Aug 11 '15 at 14:20
• Yes, it is correct – 3SAT Aug 11 '15 at 14:29

As I mentioned in the comment, you should divide the interval $[2, 4]$ to two parts, \begin{align*}S = &\int_2^3\left(\frac{2}{x - 1} - 1\right) dx \;{\color{red} -}\int_3^4\left(\frac{2}{x - 1} - 1\right)dx \\ = & \left[2\log(3 - 1) - 2\log(2 - 1) - 1\right] - \left[2\log(4 - 1) - 2\log(3 - 1) - 1\right] \\ = & 4\log 2 - 2\log 3. \end{align*}

Essentially finding the area under in the curve is integrating. So in short, yes, that would solve the question.