0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$$2x+\frac2x=5$$

$$2x^2-5x+2=0$$

$$x=\frac{5\pm3}4=\frac12,2$$

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

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.