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