I'm teaching myself definite integration and found a question like this. I have to sketch the region and then find the area through definite integration. Can anyone help? The following is the question.

Sketch the region enclosed by the line $2x + 2y = 5$ and the curve $y = 1/x$. Evaluate the area of the region with a definite integral.

  • $\begingroup$ Sorry I edited the line there now, they should cross. $\endgroup$ – Shauna Feb 28 '17 at 12:42
  • $\begingroup$ I googled the two graphs and they crossed close to 0 on the x axis $\endgroup$ – Shauna Feb 28 '17 at 12:45
  • $\begingroup$ Sorry, that was my bad. $\endgroup$ – Simply Beautiful Art Feb 28 '17 at 12:46

The first step for these types of problems is to solve for your bounds of integration. This is usually done by substiting in an equation solved for $y$ into the other.




We then decide which graph is above the other, since we take our integrals as follows:

$$\int_{1/2}^2\text{top - bottom}\ dx$$

We must also convert $2x+2y=5$ into $y=\frac52-x$. Finally, we have

$$\int_{1/2}^2\left(\frac52-x\right)-\frac1x\ dx=\frac52x-\frac12x^2-\ln|x|\bigg|_{1/2}^2\\\boxed{=\frac{15}8-\ln(4)}$$

  • $\begingroup$ can you explain to me how you got 2x^2−5x+2=0? I cant see how you got the 2x^2. Thanks $\endgroup$ – Shauna Feb 28 '17 at 13:24
  • $\begingroup$ I just multiplied both sides by $x$ and then moved everything to one side. $\endgroup$ – Simply Beautiful Art Feb 28 '17 at 14:50
  • $\begingroup$ @SimplyBeautifulArt I'm really sorry. I wanted to learn how to write integration with limits . And doing so I just messed your answer. Please edit it. $\endgroup$ – user187604 Jul 12 '18 at 2:10

Sketching is up to you. Here's how you find the area. First of all, convert the equation of the straight line to $y$ as a function of $x$:

$$ y = \frac52 - x. $$

Now we find the points of intersection of this function and $y = \frac1x$. Thus,

$$ \frac1x = \frac52 - x \\ \implies 2x^2 - 5x + 2 = 0 \\ \implies x = \frac{5 \pm 3}{4} = \frac12, 2 $$

The points of intersection are $(\frac12, 2)$ and $(2,\frac12)$. In the region enclosed by the two curves, $\frac52 - x > \frac1x$. Hence, we integrate their difference between them in the interval $[\frac12, 2]$ to get the area:

$$ A = \int_{0.5}^2 \left(\frac52 - x - \frac1x \right) dx \\ \implies A = \left[ \frac52 x - \frac12 x^2 - \ln x \right]_{0.5}^2. $$

The rest is up to you to calculate.

  • $\begingroup$ Can I just ask, when converting the equation, I got 5-2x/2. Where did your 2 multiplied by x go to? $\endgroup$ – Shauna Feb 28 '17 at 13:14
  • $\begingroup$ $(5 - 2x)/2 = \frac52 - x$. $\endgroup$ – Nilabro Saha Feb 28 '17 at 13:16

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