Integral substitution for a polynomial function

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    Integral Substitution
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Discussion Overview

The discussion revolves around the process of integral substitution for the polynomial function $$\int_{-1}^{1} \left(x^4-2x\right)^6\left(2x^3-1\right)\,dx$$. Participants explore the implications of changing variables and the necessity of adjusting integration limits accordingly.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant proposes the substitution $u=x^4-2x$ and derives $du=4x^3-2 dx =2(2 x^3-1) dx$, leading to the integral $$\frac{1}{2}\int_{-1}^{1} u^6\,du$$.
  • Another participant points out that the limits of integration need to be changed when performing the substitution and asks for the values of $u$ at the original limits $x = -1$ and $x = 1$.
  • A later reply confirms that the substitution is correct but emphasizes the need to adjust the limits, calculating $u$ at the endpoints to find $u = 3$ when $x = -1$ and $u = -1$ when $x = 1$.
  • Another participant notes that this leads to the equivalent expression $$-\dfrac{1}{2}\int_{-1}^3u^6\,du$$.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the substitution but disagree on the handling of the limits of integration, with some emphasizing the need for adjustment while others initially overlook this aspect.

Contextual Notes

The discussion highlights the importance of changing limits in integral substitution, but does not resolve the implications of this change on the evaluation of the integral.

karush
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W.8.1.18

$$\int_{-1}^{1} \left(x^4-2x\right)^6\left(2x^3-1\right)\,dx $$

$u=x^4-2x$ $du=4x^3-2 dx =2(2 x^3-1) dx$

$$\frac{1}{2}\int_{-1}^{1} u^6\,du $$

Is this ok
 
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karush said:
W.8.1.18

$$\int_{-1}^{1} \left(x^4-2x\right)^6\left(2x^3-1\right)\,dx $$

$u=x^4-2x$ $du=4x^3-2 dx =2(2 x^3-1) dx$

$$\frac{1}{2}\int_{-1}^{1} u^6\,du $$

Is this ok
You forgot to change the limits.

If $u = x^4-2x$ then what's $u$ when $x = -1$ and when $x = 1$?
 
not sure what you mean, should $u$ be something else?
 
karush said:
not sure what you mean, should $u$ be something else?
No, what you have done is correct. However, the limits should have changed. You're integrating the first integral between $x=-1$ and $x=1$. When you make the sub $u = x^4-2x $ your limits change! So $u = x^4-2x$ and when $x = -1$ you have $u = (-1)^4-2(-1) = 1+2 = 3$ and when $x = 1$ you have $u = (1)^4-2(1) = 1-2 = -1$. So you should have got:$$\displaystyle \frac{1}{2} \int_3^{-1} u^6 \,du$$
 
... which is equivalent to

$$-\dfrac12\int_{-1}^3u^6\,du$$.
 

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