Integration Q (1 Viewer)

turnerloos · Oct 1, 2015

Could someone explain how they integrated that?

Screen Shot 2015-10-01 at 12.08.29 am.png

InteGrand · Oct 1, 2015

turnerloos said:
Could someone explain how they integrated that?View attachment 32476

$It's the reverse chain rule, but here it is with explicit substitution:$

$Given: $v=\int x \left(1-\frac{4}{\pi^2}x^2 \right)^n\text{ d}x=\int \left(1-\frac{4}{\pi^2}x^2 \right)^n \cdot x\text{ d}x$.$

$Substitute $z=1-\frac{4}{\pi^2}x^2$, so $\mathrm{d}z = -\frac{8}{\pi^2}x\text{ d}x\Rightarrow x\text{ d}x = -\frac{\pi^2}{8}\text{ d}z$. Implementing these substitutions, the integral is$

$v=\int z^n \cdot \left(-\frac{\pi^2}{8}\right)\text{ d}z$

$$=-\frac{\pi^2}{8}\times \frac{z^{n+1}}{n+1}$ (ignore the $+C$)$

$$=-\frac{\pi^2}{8}\times \frac{\left( 1-\frac{4}{\pi^2}x^2 \right)^{n+1}}{n+1}$, which is what they had.$

sy37 · Oct 1, 2015

if v = (-pi)^2 / 8(n+1) * (1 - 4x^2 / pi^2)

then dv/dx = x(1 - 4x^2 / pi^2)^nth by differentiation

Not sure how to format it, I usually use MathJax

It's 3:26 am excuse any errors, if any..

turnerloos · Oct 1, 2015

^can someone report him? Sick of him trolling.

Drsoccerball · Oct 1, 2015

turnerloos said:
^can someone report him? Sick of him trolling.

Arnt you the one who always posts links of tafe enrolment :L...

Squar3root · Oct 1, 2015

turnerloos said:
^can someone report him? Sick of him trolling.

you can do it?

Drsoccerball said:
Arnt you the one who always posts links of tafe enrolment :L...

lel

turnerloos · Oct 1, 2015

Drsoccerball said:
Arnt you the one who always posts links of tafe enrolment :L...

I'm providing an alternate pathway for talented students like sy37.

Integration Q (1 Viewer)

turnerloos

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InteGrand

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sy37

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turnerloos

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Drsoccerball

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Squar3root

realest nigga

turnerloos

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