sorry, another one @solomonzelman
It's ok:)
I'll have to use approimation with rectangles (MRAM) right?
What does MRAM stand for?
Rectangular approximation method, m means the hieght of the rectnagles is determined by the midpoint
of subsections
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
illustration: https://images.duckduckgo.com/iu/?u=http%3A%2F%2Fi1.ytimg.com%2Fvi%2FF045BKT-vPs%2Fmaxresdefault.jpg&f=1
M - midpoint
the first M
(btw, the greatest precision, and errror twice smaller than off trapezoidal)
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?
Wait, wouldn't it be 0?
wouldn't what be 0?
the area of the region, and thus the average value
um....i don't think so? why do you think 0?
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 \(-\varphi\)
because the region is symmetric about the origin.
you can see this very well on the graph too.... https://www.desmos.com/calculator/dwlqhoaq6h
area can't be - though....
yes it could be
The area could be negative when it is below the x-axis. We are in the calculus land, not geometry land.
So, the positive and negative areas (which have the same magnitude) will cancel each other out, and your area is going to be 0.
ah....
so the answer is 0 then
yes.
Awesome!
And there is another way to determine this
how?
|dw:1452208374435:dw|
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] )
So the average is zero.
And my hypothesis can be shown.
\(\color{#000000 }{ \displaystyle \tan(-z)=-\tan(z) }\)
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
Anyway ... way 1 would suffice too, I guess I am being too complicated
that makes sense. thank you. :)
yw
Join our real-time social learning platform and learn together with your friends!