Find the area of the shaded region shown in the first quadrant between f ( x ) = 1 x f\left(x\right)=\frac{1}{x} & g ( x ) = − 1 4 x + 5 4 . g\left(x\right)=-\frac14x+\frac54.

this problem, we're asked to find the area of the shaded region shown in the first quadrant between the functions f of x equals one over x and g of x equals negative one fourth x plus five fourths. So let's go ahead and get started here. We know that we need to set up a definite integral where we take our top function minus our bottom function, but we're not explicitly given the bounds of this integral. So that means that we need to find them by finding the points where our functions intersect each other, which we can do by setting our functions equal to each other. So if I take one over x and set that equal to negative one fourth x plus five over four, I then wanna solve for x here. Now there's a couple of things that I wanna do to rearrange this equation and make this much easier to solve. The first thing I'm gonna do here is multiply both sides by x. So in doing that, I'll cancel this x out over here on this left side, leaving me with one equals negative one fourth x squared plus five fourths x. Now I wanna go ahead and subtract one from both sides to get everything all on one side. So then zero is equal to negative one fourth x squared plus five fourths x minus one. Now dealing with fractions in a quadratic equation is not ideal, but we can get rid of these fractions by multiplying by four. Now in doing this, I'm actually gonna multiply by negative four because I also just wanna go ahead and get rid of this negative to make this a bit easier to factor. So if we multiply everything here by negative four, I'm gonna do the same thing to this side, but that's not gonna change anything. So this is still zero, and then this is equal to positive x squared minus five x and then plus four. So now I have this much easier quadratic equation to work with. So I can go ahead and just factor this in order to find out what these intersection points are. So x squared minus five x plus four. In order to factor this, I'm looking for two values that multiply to four and add to negative five. What are those values? Well, it's going to be negative one and negative four. So this factors here to x minus one times x minus four, and that's equal to zero. Now since this is equal to zero, I can take each of these individual factors and set them equal to zero to find these intersection points. So x minus one equals zero and x minus four equals zero. Now if I add one to both sides here, this gives me that x is equal to positive one and then doing the same thing over here adding four to both sides, this tells me that x is equal to positive four. So these are my two intersection points and thus the bounds of this integral. So let's go ahead and set up this area integral. Going ahead and blocking off that work here. My area is going to be equal to the definite integral from one to four based on those two values that we just found of our top function. And coming back over to our graph here, I can see that the function on top is that line g of x. So that is negative one fourth x plus five over four, and then minus our bottom function here that is one over x and then integrating with respect to x. So we can go ahead and find this integral negative one fourth x. If I go ahead and use the power rule here, I'm adding one to that exponent that makes it two and then dividing by that two. So this makes this negative one eighth x squared and then plus five over four times x and then minus the natural log of x bounded from one to four. So now we just need to plug those bounds in and I'm going to go ahead and come down here to give myself a bit more space in doing so. So plugging in that upper bound, this is negative one over eight times four squared plus five over four times four minus the natural log of four. So that's that upper bound plugged in. Then we're subtracting off that lower bound plugged in. So that's negative one eighth. One squared plus five over four times one minus the natural log of one. So I'm going to go ahead and go through all of the simplification steps. But at this point, if you wanna just go ahead and plug this into your calculator, if that if that is something that your professor expects you to do on an exam or on your homework, you can go ahead and skip all the way to the end and check your answer. But if you're fully doing this by hand at this point, it would be good to go through all of these algebra steps with me in case you made an error along the way. So let's go ahead and talk through all of this algebra. Now looking through this, I can see a couple of different things happening here. Looking at this first set of terms here with that upper bound plugged in, these fours are going to cancel out there. And then I have this four squared that is 16. So that's negative 16 over eight. Negative 16 over eight. I can reduce that fraction there. So that's negative two. So if I have negative two plus five minus the natural log of four. So doing a bit of simplification there, I can also go ahead and combine those like terms. Negative two plus five. That's going to give me positive three. So this is really three minus the natural log of four. So I'm gonna keep that as is for now and then start to look at this next set of terms here. So here I have negative one eighth times one squared. So that's gonna stay as negative one eighth because one squared is just one. And then five fourths times one, that is also going to stay, so plus five fourths. And then natural log of one is just zero, so I can go ahead and cancel out that term and not worry about it anymore. Now looking at these two terms, they are both fractions but I want them to be able to be added together. So I'm gonna go ahead and give that five fourths a common denominator with that negative one eighth. If I go ahead and multiply both the numerator and denominator by two, that is 10 over eight. So this really becomes nine over eight if I take that 10 over eight and subtract that one eighth. So everything that I'm left with here, I have three minus the natural log of four and then minus nine over eight. Now we want to go ahead and combine these like terms because they're both constants, which I can go ahead and do by giving that three a common denominator here, multiplying it by eight over eight, which is going to be 24 over eight, and then having that nine over eight there and then minus the natural log of four. Now, since those have a common denominator, I can go ahead and combine those like terms again as well. So that gives me 15 over eight minus the natural log of four. Now if you do not use a calculator in your course, this would be your final answer here. But if you wanna go ahead and plug this into your calculator in order to get a decimal value here with that natural log of four, this will end up giving you an answer of 0.489 for the area between these two functions that we see in that first quadrant there. Now if you have any questions about this final answer or the work that we did to get there feel free to let us know in the comments. I'll see you in the next video.

9. Graphical Applications of Integrals

Area Between Curves

Business Calculus

9. Graphical Applications of Integrals

Area Between Curves

Struggling with Business Calculus?

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Multiple Choice

Find the area of the shaded region shown in the first quadrant between $f (x) = \frac{1}{x}$ & $g (x) = - \frac{1}{4} x + \frac{5}{4} .$

4.239

1.386

0.489

Verified step by step guidance

Identify the functions that define the boundaries of the shaded region: f(x) = $\frac{1}{x}$ and g(x) = -$\frac{1}{4}$x + $\frac{5}{4}$.

Determine the points of intersection between the two functions by setting f(x) equal to g(x) and solving for x: $\frac{1}{x}$ = -$\frac{1}{4}$x + $\frac{5}{4}$.

Solve the equation $\frac{1}{x}$ = -$\frac{1}{4}$x + $\frac{5}{4}$ to find the x-values where the curves intersect. This involves finding a common denominator and solving the resulting quadratic equation.

Once the points of intersection are found, set up the integral to find the area between the curves. The area A is given by the integral from the lower intersection point to the upper intersection point of (f(x) - g(x)) dx.

Evaluate the integral $\int$_{a}^{b} $\left$($\frac{1}{x}$ - $\left$(-$\frac{1}{4}$x + $\frac{5}{4}$$\right$)$\right$) dx, where a and b are the x-values of the intersection points, to find the area of the shaded region.

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Struggling with Business Calculus?

Find the area of the shaded region shown in the first quadrant between f(x)=1xf\(\left\)(x\(\right\))=\(\frac{1}{x}\) & g(x)=−14x+54.g\(\left\)(x\(\right\))=-\(\frac\)14x+\(\frac\)54.

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Area Between Curves practice set

Find the area of the shaded region shown in the first quadrant between $f (x) = \frac{1}{x}$ & $g (x) = - \frac{1}{4} x + \frac{5}{4} .$