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)
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$\begingroup$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: $$\int_0^1 \left(x - \frac x4\right)\,dx \quad + \quad \int_1^2 \left(\frac 1x - \frac x4\right)\,dx $$
$\endgroup$ 10 $\begingroup$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: $$ \left[\int_0^1 x - \frac x4~dx \right] + \left[\int_1^2 \frac 1x - \frac x4~dx \right] $$
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$Make a scetch of all three curves over the domain $x>0$. Can you see how they form and bound a triangle-like figure?
$\endgroup$ $\begingroup$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.
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