# sketch the region enclosed by the given curves and find its area

sketch the region enclosed by the given curves and find its area:

$$y=\frac 1x,\; y=x,\; y=\frac x4,\; x>0.$$

I have no problem sketching the area between the curves but there are three, and only one constant value, so I don't know what to put as my second a/b value. I tried using 1/x as a b value but that just gave me an equation answer and the answer isn't an equation.

edit: i forgot about the intersections as constant values. but now I see how to split them up.

This is chapter 6.1 calculus James Stewart btw (area between curves)

• Just FYI: there are many calculus textbooks each with their own chapter/section scheme. "Chapter 6.1" isn't useful information without telling us what textbook you're using. – Adam Saltz Sep 2 '13 at 21:28

From Wolfram Alpha, we can sketch the curves to find the area of interest:

Note that we need to find the points of intersection: at $x = 0$ the lines $y = x, \;y = \frac x{4}$ intersect. At $x= 1$, the lines $y = x$ and $y = \frac 1x$ intersect. At $x = 2,$ the lines $y = \frac 1x$ and $y = \frac x4$ intersect. You can solve this by integrating between the relevant curves from $x = 0$ to $x = 1$, and likewise integrating between the relevant curves between $x = 1$ and $x = 2$, then summing: $$\int_0^1 \left(x - \frac x4\right)\,dx \quad + \quad \int_1^2 \left(\frac 1x - \frac x4\right)\,dx$$

• ok I solved it but I got ln2 - 1/2 and the answer is ln2 O_o i double checked calculations what did i do wrong?? – J L Sep 2 '13 at 21:39
• You have to evaluate the second integral $F(2) - F(1)$ – Namaste Sep 2 '13 at 21:42
• First integral $\frac 12 - \frac 18 = \frac 38$. Second integral: $\ln 2 - \frac 12 - (\ln 1 - \frac 18) = \ln 2 - \frac 38$. Sum, we get $\ln 2$. – Namaste Sep 2 '13 at 21:44
• i figured it out i wrote 7/8 instead of 3/8 for some reason – J L Sep 2 '13 at 21:45
• Good then, that should agree. Is all good? – Namaste Sep 2 '13 at 21:47

Hint: Suppose that we integrate with respect to $x$. Since there are two types of upper curves, draw a vertical line at $x=1$ (where the two upper curves of $y=x$ and $y=1/x$ intersect) that splits the region into two cases. You should obtain: $$\left[\int_0^1 x - \frac x4~dx \right] + \left[\int_1^2 \frac 1x - \frac x4~dx \right]$$

• ohhh yeahh i see. thanks for your help :) I forgot about spliting things into two integrals – J L Sep 2 '13 at 21:26

If you sketched the region correctly then you should be seeing a sort of triangle whose base is given by the line $y=x/4$ whose left side is given by $y=x$ and whose right side is given my $y=1/x$.

The first thing you have to do is to identify the crossing points, that is, the endpoints of this distorted triangle. One crossing point is obviously given by $(0,0)$ where the lines $y=x$ and $y=x/4$ meet, for the next one you have to solve the equation $$\frac{1}{x}=x$$ from where you obtain $x=1$ (we care only about the positive solution since $x>0$). The other vertex of the desired region is given by solving $$\frac{1}{x}=\frac{x}{4}$$ from where you can obtain the solution $x=2$. Look at your sketched region, we will integrate in terms of x-slices. Usually one integrates the area between the two curves $f(x)$ and $g(x)$ in the $x$-region $[a,b]$ as $\int_a^b f(x)-g(x) \; dx$. In this case however, the expression for the upper limit changes exactly at $x=1$ so there should be TWO subtractions instead of one. The desired expression for the area $A$ is $$A = \int_{0}^1 x-\frac{x}{4}\; dx + \int_{1}^2 \frac{1}{x}-\frac{x}{4} \; dx$$ And you should be able to calculate that integral yourself. Hope that helps.

Make a scetch of all three curves over the domain $x>0$. Can you see how they form and bound a triangle-like figure?

What do you mean by "only one constant value"? If you sketch the three curves, you'll see two regions which look like triangles with one bent edge (the $y=1/x$ part). In one of those regions, all the points have positive $x$-coordinates. In the other, the $x$-coordinates are all negative. Integrate to find the area of the positive one.