I have no difficulty sketching the area between the curves yet there room three, and only one continuous value, so i don"t recognize what to placed as my second a/b value. Ns tried making use of 1/x together a b value yet that just offered me one equation answer and the answer isn"t an equation.

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edit: ns forgot about the intersections as constant values. Yet now ns see just how to split them up.

This is thing 6.1 calculus James Stewart btw (area between curves)  From Wolfram Alpha, we deserve to sketch the curve to uncover the area the interest: Note that we need to find the points of intersection: in ~ $x = 0$ the currently $y = x, \;y = \frac x4$ intersect. At $x= 1$, the currently $y = x$ and also $y = \frac 1x$ intersect. At $x = 2,$ the currently $y = \frac 1x$ and $y = \frac x4$ intersect. You have the right to solve this through integrating in between the relevant curves native $x = 0$ come $x = 1$, and an in similar way integrating between the appropriate curves between $x = 1$ and $x = 2$, then summing: $$\int_0^1 \left(x - \frac x4\right)\,dx \quad + \quad \int_1^2 \left(\frac 1x - \frac x4\right)\,dx$$ Hint: suppose that we combine with respect to $x$. Because there space two types of top curves, attract a vertical line at $x=1$ (where the two upper curve of $y=x$ and $y=1/x$ intersect) the splits the an ar into two cases. You must obtain:$$\left<\int_0^1 x - \frac x4~dx \right> + \left<\int_1^2 \frac 1x - \frac x4~dx \right>$$ If girlfriend sketched the region correctly then you must be seeing a type of triangle whose base is provided by the heat $y=x/4$ who left next is given by $y=x$ and whose appropriate side is given my $y=1/x$.

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The very first thing you need to do is to recognize the cross points, that is, the endpoints of this distorted triangle. One crossing allude is obviously given by $(0,0)$ where the present $y=x$ and also $y=x/4$ meet, because that the following one you have to solve the equation $$\frac1x=x$$from whereby you achieve $x=1$ (we treatment only about the confident solution because $x>0$). The other vertex of the desired an ar is given by resolving $$\frac1x=\fracx4$$from wherein you can attain the equipment $x=2$. Watch at your sketched region, we will combine in terms of x-slices. Typically one integrates the area between the 2 curves $f(x)$ and $g(x)$ in the $x$-region $$together \int_a^b f(x)-g(x) \; dx. In this instance however, the expression for the upper limit transforms exactly in ~ x=1 so there have to be 2 subtractions instead of one. The wanted expression for the area A is$$ A = \int_0^1 x-\fracx4\; dx + \int_1^2 \frac1x-\fracx4 \; dx And friend should be able to calculate the integral yourself. Hope the helps.