Integral equation. Please check

  • Thread starter yyttr2
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  • #1
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I have recently been playing with integrals and I still do not fully understand them.

Usually the best way for me to learn is to play with values and figure it out on my own, so I would like you (physics forum) to check my work so far.

After a bit of time I got this:

[tex]\int_{0}^{t}ax^ndx=\frac{at^{n+1}}{n+1}, n\neq-1[/tex]


and after figuring out:
[tex]\int_{0}^{t}ax+bxdx=\int_{0}^{t}axdx+\int_{0}^{t}bxdx[/tex]

I got this equation.

[tex]\int_{0}^{t}ax^n+bx^{n-1}+cx^{n-2} \cdots +ex dx=\int_{0}^{t}ax^ndx+\int_{0}^{t}bx^{n-1}dx+\int_{0}^{t}cx^{n-2}dx \cdots +\int_{0}^{t}(ex)dx=\frac{at^{n+1}}{n+1}+\frac{bt^n}{n}+\frac{ct^n-1} {n-1} \cdots +et[/tex]

Is this true for all polynomials (in standard form)? or does it even work at all?
 
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Answers and Replies

  • #2
tiny-tim
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Hi yyttr2! :wink:

Yes, that's correct.

The integral of the sum is the sum of the integrals …

you can always split a sum (any sum, not just polynomials) into its separate parts, and integrate them separately. :smile:

(unless, of course, that requires you to add an ∞ and a minus ∞)
 
  • #3
HallsofIvy
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You can prove, in general, that
[tex]\int (f_1(x)+ f_2(x)+ \cdot\cdot\cdot+ f_n(x))dx= \int f_1(x)dx+ \int f_2(x)dx+ \cdot\cdot\cdot+ \int f_n(x) dx[/tex]
by induction on n.
 

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