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Revision as of 21:10, 26 August 2008 editWoland37 (talk | contribs)Pending changes reviewers4,618 editsm Requesting speedy deletion (CSD A1). (TW)← Previous edit Revision as of 21:21, 26 August 2008 edit undoLindySoul (talk | contribs)230 edits ← Created page with 'Who the hell are you and why would you be so bigoted and stupid to delete this so quickly? I just started writing it. Isnt wikipedia supposed to be a continuous ...'Next edit →
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Who the hell are you and why would you be so bigoted and stupid to delete this so quickly?
{{db-nocontext}}

I just started writing it.

Isnt wikipedia supposed to be a continuous work-in-progress? Isnt the point of this site for other people to contribute at will?

Who the hell are you to delete perfectly legit content capable of helping people?

The least you could do is give me a full five minutes to finish writing the article before judging it.

As for renaming it, I dont mind. That is fine. But where would you suggest I put it?

== Integration Using Parametric Derivatives == == Integration Using Parametric Derivatives ==


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<math>\int_0^{\infty} x^2 e^{-3x} dx = \frac{2}{3^3} = \frac{2}{27}</math> <math>\int_0^{\infty} x^2 e^{-3x} dx = \frac{2}{3^3} = \frac{2}{27}</math>

== Integration Using Complex Analysis ==

Suppose we wanted to integrate:

<math>\int e^x \cos x dx</math>

Instead of using ], we may substitute the cosine function for its Euler form: <math>\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}</math>

<math>\int e^x \cdot \frac{e^{ix} + e^{-ix}}{2} dx</math>

<math>1\over 2 \int e^{x(1+i)} + e^{x(1-i)} dx</math>

Revision as of 21:21, 26 August 2008

Who the hell are you and why would you be so bigoted and stupid to delete this so quickly?

I just started writing it.

Isnt wikipedia supposed to be a continuous work-in-progress? Isnt the point of this site for other people to contribute at will?

Who the hell are you to delete perfectly legit content capable of helping people?

The least you could do is give me a full five minutes to finish writing the article before judging it.

As for renaming it, I dont mind. That is fine. But where would you suggest I put it?

Integration Using Parametric Derivatives

Suppose you wanted to find the integral:

∫ 0 ∞ x 2 e − 3 x d x {\displaystyle \int _{0}^{\infty }x^{2}e^{-3x}dx}

We may solve this by starting with the integral:

∫ 0 ∞ e − t x d x = {\displaystyle \int _{0}^{\infty }e^{-tx}dx=}

[ e − t x − t ] 0 ∞ = {\displaystyle \left_{0}^{\infty }=}

[ lim x → ∞ e − t x − t ] − [ e − t 0 − t ] = {\displaystyle \left-\left=}

[ 0 ] − [ 1 − t ] = 1 t {\displaystyle \left-\left={\frac {1}{t}}}

Now that we know:

∫ 0 ∞ e − t x d x = 1 t {\displaystyle \int _{0}^{\infty }e^{-tx}dx={\frac {1}{t}}}

Suppose we found the second derivative with respect, not to x, but to t:

d 2 d t 2 ∫ 0 ∞ e − t x d x = d 2 d t 2 1 t {\displaystyle {\frac {d^{2}}{dt^{2}}}\int _{0}^{\infty }e^{-tx}dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}}

∫ 0 ∞ d 2 d t 2 e − t x d x = d 2 d t 2 1 t {\displaystyle \int _{0}^{\infty }{\frac {d^{2}}{dt^{2}}}e^{-tx}dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}}

∫ 0 ∞ d d t − x e − t x d x = d d t − 1 t 2 {\displaystyle \int _{0}^{\infty }{\frac {d}{dt}}-xe^{-tx}dx={\frac {d}{dt}}-{\frac {1}{t^{2}}}}

∫ 0 ∞ x 2 e − t x d x = 2 t 3 {\displaystyle \int _{0}^{\infty }x^{2}e^{-tx}dx={\frac {2}{t^{3}}}}

Now notice that this solution takes the same form as the original proposed question. In the original problem, t = 3. Substituting that into our new solution equation:

∫ 0 ∞ x 2 e − 3 x d x = 2 3 3 = 2 27 {\displaystyle \int _{0}^{\infty }x^{2}e^{-3x}dx={\frac {2}{3^{3}}}={\frac {2}{27}}}

Integration Using Complex Analysis

Suppose we wanted to integrate:

∫ e x cos ⁡ x d x {\displaystyle \int e^{x}\cos xdx}

Instead of using Integration by parts, we may substitute the cosine function for its Euler form: cos ⁡ θ = e i θ + e − i θ 2 {\displaystyle \cos \theta ={\frac {e^{i\theta }+e^{-i\theta }}{2}}}

∫ e x ⋅ e i x + e − i x 2 d x {\displaystyle \int e^{x}\cdot {\frac {e^{ix}+e^{-ix}}{2}}dx}

1 2 ∫ e x ( 1 + i ) + e x ( 1 − i ) d x {\displaystyle 1 \over 2\int e^{x(1+i)}+e^{x(1-i)}dx}