Finding area (application of definite integral)

[Noah1]

Hi, I am stuck on this question and was wondering if anyone could help me. The topic is integral equations.

A block of land is bounded by two fences running North-South 5 km apart a fence line which is approximated by the function N=0.5E and a road which is approximated by the curve N=0.25E^2-E+30. N and E are measured in kilometres. Calculate the area of this block of land.

 

[Evgeny.Makarov]

Re: Stuck on these questions

Do you know that the area under the graph of $f(x)$ between $x_1$ and $x_2$ is $\displaystyle\int_{x_1}^{x_2}f(t)\,dt$?

 

[Noah1]

Re: Stuck on these questions

Do you know that the area under the graph of $f(x)$ between $x_1$ and $x_2$ is $\displaystyle\int_{x_1}^{x_2}f(t)\,dt$?

no what is that

 

[Evgeny.Makarov]

Re: Stuck on these questions

I guess this can serve as the definition of the (net) area under the curve. See, for example, here and here or search for "area under the curve".

In general, it's a very bad idea to attempt solving exercises without learning the relevant theory. Doing so is simply impossible. I believe this problem is given to you as an application of some lesson about integrals, and it would be absurd to ignore the lesson and go straight for problems.

 

[Noah1]

Re: Stuck on these questions

I guess this can serve as the definition of the (net) area under the curve. See, for example, here and here or search for "area under the curve".

In general, it's a very bad idea to attempt solving exercises without learning the relevant theory. Doing so is simply impossible. I believe this problem is given to you as an application of some lesson about integrals, and it would be absurd to ignore the lesson and go straight for problems.

I have tried to solve it that way do I inset both of the equations like this ∫_0^5▒〖0.25E^2-E-0.5E+30〗 dE

 

[Noah1]

Re: Stuck on these questions

I have tried to solve it that way do I inset both of the equations like this ∫_0^5▒〖0.25E^2-E-0.5E+30〗 dE

is the answer 141.67?

 

[MarkFL]

We are given a road which follows:

$$R=\frac{1}{4}E^2-E+30$$

and a fence which follows:

$$F=\frac{1}{2}E$$

I would first see if the road and the fence intersect anwhere:

$$\frac{1}{4}E^2-E+30=\frac{1}{2}E$$

$$E^2-4E+120=2E$$

$$E^2-6E+120=0$$

We see the discriminant is negative, thus the two functions do not intersect, and we see that:

$$R(0)>F(0)$$

And so we may conclude that for all $E$, we must have:

$$R>F$$

Now, we aren't told where the two other fences (the two running north/south) lie (only that they are 5 km apart), so I would put them at:

$$E_1=a$$

$$E_2=a+5$$

And so the area $A$ (in $\text{km}^2$) bounded by the road and 3 fences is given by:

$$A=\int_{E_1}^{E_2} R-F\,dE=\int_{a}^{a+5} \left(\frac{1}{4}E^2-E+30\right)-\left(\frac{1}{2}E\right)\,dE=\frac{1}{4}\int_{a}^{a+5} E^2-6E+120\,dE$$

Apply the FTOC:

$$A=\frac{1}{4}\left[\frac{1}{3}E^3-3E^2+120E\right]_{a}^{a+5}=\frac{1}{12}\left(\left((a+5)^3-9(a+5)^2+360(a+5)\right)-\left(a^3-9a^2+360a\right)\right)=\frac{5}{12}\left(3a^2-3a+340\right)$$

Now, if we assume that $a=0$, then we would have:

$$A=\frac{425}{3}=141.\overline{6}$$

 

[Noah1]

We are given a road which follows:

$$R=\frac{1}{4}E^2-E+30$$

and a fence which follows:

$$F=\frac{1}{2}E$$

I would first see if the road and the fence intersect anwhere:

$$\frac{1}{4}E^2-E+30=\frac{1}{2}E$$

$$E^2-4E+120=2E$$

$$E^2-6E+120=0$$

We see the discriminant is zero, thus the two functions do not intersect, and we see that:

$$R(0)>F(0)$$

And so we may conclude that for all $E$, we must have:

$$R>F$$

Now, we aren't told where the two other fences (the two running north/south) lie (only that they are 5 km apart), so I would put them at:

$$E_1=a$$

$$E_2=a+5$$

And so the area $A$ (in $\text{km}^2$) bounded by the road and 3 fences is given by:

$$A=\int_{E_1}^{E_2} R-F\,dE=\int_{a}^{a+5} \left(\frac{1}{4}E^2-E+30\right)-\left(\frac{1}{2}E\right)\,dE=\frac{1}{4}\int_{a}^{a+5} E^2-6E+120\,dE$$

Apply the FTOC:

$$A=\frac{1}{4}\left[\frac{1}{3}E^3-3E^2+120E\right]_{a}^{a+5}=\frac{1}{12}\left(\left((a+5)^3-9(a+5)^2+360(a+5)\right)-\left(a^3-9a^2+360a\right)\right)=\frac{5}{12}\left(3a^2-3a+340\right)$$

Now, if we assume that $a=0$, then we would have:

$$A=\frac{425}{3}=141.\overline{6}$$

Thanks for that is this question the same principal?
Find the Physical area between that curves y=sin(x) and y=cos(x) from x=pi/2 to x=3pi/2

 

[MarkFL]

Thanks for that is this question the same principal?
Find the Physical area between that curves y=sin(x) and y=cos(x) from x=pi/2 to x=3pi/2

Yes, you will need to determine any intersections of the two functions on the given interval, and then determine which function is on top in each sub-interval, and within each sub-interval make your integrand the top function minus the bottom function.

I would begin by plotting both functions on the given interval to see the area in question and to help with determining the formula for the area. :D

 

[MarkFL]

To find the intersection(s) in the given interval, we may write:

$$\sin(x)=\cos(x)$$

Now, observing that $$\cos\left(\frac{\pi}{2}\right)=\cos\left(\frac{3\pi}{2}\right)=0$$ and:

$$\cos\left(\frac{\pi}{2}\right)\ne\sin\left(\frac{\pi}{2}\right)$$

$$\cos\left(\frac{3\pi}{2}\right)\ne\sin\left(\frac{3\pi}{2}\right)$$

We may then divide though by $\cos(x)$ without losing any solutions on the given interval to get:

$$\tan(x)=1$$

And, given the periodicity of the tangent function and the quadrant I solution of $$x=\frac{\pi}{4}$$, we may give the general solution as:

$$x=\frac{\pi}{4}+k\pi=\frac{\pi}{4}(4k+1)$$ where $k\in\mathbb{Z}$

Now, along with the given interval and what we found at the endpoints, we may write:

$$\frac{\pi}{2}<\frac{\pi}{4}(4k+1)<\frac{3\pi}{2}$$

$$2<4k+1<6$$

$$1<4k<5$$

$$\frac{1}{4}<k<\frac{5}{4}$$

Hence (given that $k$ is an integer):

$k=1$

And so the intersection of the two functions occurs for:

$$x=\frac{\pi}{4}(4(1)+1)=\frac{5\pi}{4}$$

Now, observing that:

$$\sin\left(\frac{\pi}{2}\right)>\cos\left(\frac{\pi}{2}\right)$$

$$\sin\left(\frac{3\pi}{2}\right)<\cos\left(\frac{3\pi}{2}\right)$$

We may conclude:

$$\sin(x)>\cos(x)$$ on $$\left[\frac{\pi}{2},\frac{5\pi}{4}\right)$$

$$\cos(x)>\sin(x)$$ on $$\left(\frac{5\pi}{4},\frac{3\pi}{2}\right]$$

And so the area $A$ in question will be given by:

$$A=\int_{\frac{\pi}{2}}^{\frac{5\pi}{4}} \sin(x)-\cos(x)\,dx+\int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} \cos(x)-\sin(x)\,dx=(1+\sqrt{2})+(\sqrt{2}-1)=2\sqrt{2}$$