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)


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From Wolfram Alpha, we deserve to sketch the curve to uncover the area the interest:

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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: $$\int_0^1 \left(x - \frac x4\right)\,dx \quad + \quad \int_1^2 \left(\frac 1x - \frac x4\right)\,dx $$


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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:$$\left<\int_0^1 x - \frac x4~dx \right> + \left<\int_1^2 \frac 1x - \frac x4~dx \right>$$


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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.