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Mathematics 18 Online
OpenStudy (aryana_maria2323):

Please help? Will FAN AND MEDAL! Find the range of the function f of x equals the integral from negative 6 to x of the square root of the quantity 36 minus t squared dt. a. [-6,0] b. [0,6] c. [0,9pie] d. [0,18pie]

OpenStudy (aryana_maria2323):

Can you please help? @zepdrix @Directrix @Nnesha @agent0smith

OpenStudy (agent0smith):

You gotta take screenshots or learn to use latex.

OpenStudy (aryana_maria2323):

what is latex?

OpenStudy (agent0smith):

The equation writing on here, press the equation button and start learning. Screenshots too. Don't write integrals in words, good god no.

OpenStudy (aryana_maria2323):

\[f(x)=\int\limits_{-6}^{x}\sqrt{36-t^2}dt\]

OpenStudy (agent0smith):

I'd start by drawing the function inside the integral. The sqrt(36-t^2). Hopefully you recognize it as a semicircle

OpenStudy (agent0smith):

Remember that the integral means you're finding the area under the curve. If you drew out that semicircle, what is the area of it?

OpenStudy (aryana_maria2323):

\[\sqrt{-(t+6)(t-6)}\] I put the original equation in and got this

OpenStudy (aryana_maria2323):

@agent0smith

OpenStudy (agent0smith):

That has nothing to do with what i said though...

OpenStudy (aryana_maria2323):

I dont understand how to draw the semi-circle, or how to find the area? That is why I put that into my calculator because that is what I thought you were suppose to do.

OpenStudy (agent0smith):

Draw means graph.

OpenStudy (aryana_maria2323):

okay so graph the equation

OpenStudy (agent0smith):

Once you realize it's a semicircle and find the radius, finding the area is easy.

OpenStudy (agent0smith):

Look. Find the radius. Then find the area. https://www.google.com/search?q=y%3Dsqrt(36-x%5E2)&oq=y%3Dsqrt(36-x%5E2)&aqs=chrome..69i57&sourceid=chrome&ie=UTF-8

OpenStudy (aryana_maria2323):

Okay the radius is either -5.4 or 5.4

OpenStudy (aryana_maria2323):

Sorry -5.66 or 5.66

OpenStudy (agent0smith):

Look at the graph i gave.

OpenStudy (agent0smith):

And... a radius is a length.... they can't be negative.

OpenStudy (aryana_maria2323):

Okay the radius is 6

OpenStudy (aryana_maria2323):

so the area is \[36\pi \]

OpenStudy (agent0smith):

The area of a circle of radius 6 is 36pi. This is not a circle.

OpenStudy (aryana_maria2323):

I know it is a semi-circle so it half of that. So the area is 18pi.

OpenStudy (agent0smith):

So if you integrated that function from x = -6 to x = 6, that's the area of the entire semircircle If you integrated from x=-6 to x=-6, what would the area be?

OpenStudy (aryana_maria2323):

9pi.?

OpenStudy (agent0smith):

If you integrate any function from x=a to x=a, the area is always the same value.

OpenStudy (agent0smith):

\[\Large \int\limits_a^a f(t)dt= 0\]

OpenStudy (aryana_maria2323):

Okay so it would be D. [0,18pi]

OpenStudy (aryana_maria2323):

@agent0smith

OpenStudy (agent0smith):

I gave you a medal right after your last post for a reason

OpenStudy (aryana_maria2323):

Ohh it never said you did. I'm sorry.

OpenStudy (aryana_maria2323):

Thank you for the help.

OpenStudy (agent0smith):

You're welcome.

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