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Mathematics 22 Online
OpenStudy (studygurl14):

sorry, another one @solomonzelman

OpenStudy (solomonzelman):

It's ok:)

OpenStudy (studygurl14):

I'll have to use approimation with rectangles (MRAM) right?

OpenStudy (solomonzelman):

What does MRAM stand for?

OpenStudy (studygurl14):

Rectangular approximation method, m means the hieght of the rectnagles is determined by the midpoint

OpenStudy (studygurl14):

of subsections

OpenStudy (solomonzelman):

Well, I am not aware of finding exact area under tan\(\theta\), asides from integration. So if they tell you to appeal to geometry then I expect they mean to approximate the region

OpenStudy (solomonzelman):

M - midpoint

OpenStudy (solomonzelman):

the first M

OpenStudy (solomonzelman):

(btw, the greatest precision, and errror twice smaller than off trapezoidal)

OpenStudy (studygurl14):

ok, so would I just be able to find the area of half, and then multiply by 2 to find the area from -pi/4 to pi/4?

OpenStudy (solomonzelman):

Wait, wouldn't it be 0?

OpenStudy (studygurl14):

wouldn't what be 0?

OpenStudy (solomonzelman):

the area of the region, and thus the average value

OpenStudy (studygurl14):

um....i don't think so? why do you think 0?

OpenStudy (solomonzelman):

The area is zero! The area under the curve of tan\(\theta\), when x=\((0~,~\pi/4)\) is say equal to \(\varphi\) then the area under the curve of tan\(\theta\), x=\((-\pi/4~,~0)\) is going to exactly be \(-\varphi\)

OpenStudy (solomonzelman):

because the region is symmetric about the origin.

OpenStudy (solomonzelman):

you can see this very well on the graph too.... https://www.desmos.com/calculator/dwlqhoaq6h

OpenStudy (studygurl14):

area can't be - though....

OpenStudy (solomonzelman):

yes it could be

OpenStudy (solomonzelman):

The area could be negative when it is below the x-axis. We are in the calculus land, not geometry land.

OpenStudy (solomonzelman):

So, the positive and negative areas (which have the same magnitude) will cancel each other out, and your area is going to be 0.

OpenStudy (studygurl14):

ah....

OpenStudy (studygurl14):

so the answer is 0 then

OpenStudy (solomonzelman):

yes.

OpenStudy (studygurl14):

Awesome!

OpenStudy (solomonzelman):

And there is another way to determine this

OpenStudy (studygurl14):

how?

OpenStudy (solomonzelman):

|dw:1452208374435:dw|

OpenStudy (solomonzelman):

For any \(\color{#000000 }{ \displaystyle f(0+x_i) }\) that you choose, say \(\color{#000000 }{ \displaystyle f(+x_i) =\varphi }\) then, \(\color{#000000 }{ \displaystyle f(-x_i) =-\varphi }\) (at least when x=[-π/4 , π/4] )

OpenStudy (solomonzelman):

So the average is zero.

OpenStudy (solomonzelman):

And my hypothesis can be shown.

OpenStudy (solomonzelman):

\(\color{#000000 }{ \displaystyle \tan(-z)=-\tan(z) }\)

OpenStudy (solomonzelman):

Thus, \(\color{#000000 }{ \displaystyle f(x_i)+f(-x_i)=\tan(x)+\tan(-x) }\) \(\color{#000000 }{ \displaystyle \tan(x)-\tan(x) =0}\) this is true for whenever x when f is defined, and our interval is defined. and the length of the interval after zro and before zero are the same

OpenStudy (solomonzelman):

Anyway ... way 1 would suffice too, I guess I am being too complicated

OpenStudy (studygurl14):

that makes sense. thank you. :)

OpenStudy (solomonzelman):

yw

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