Solving for $k$: $\pi\int_{-1}^{1} u\left(x\right)^2 \,dx = 6\pi$

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Discussion Overview

The discussion revolves around solving for the variable \( k \) in the equation \( \pi\int_{-1}^{1} u\left(x\right)^2 \,dx = 6\pi \) where \( u\left(x\right) = \sqrt{x+1} \). Participants explore the integration process and the implications of the problem, which involves concepts from calculus related to volume and solids of revolution.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that \( \pi\int_{-1}^{1} u\left(x\right)^2 \,dx = 2\pi \) and sets up the equation \( \pi\int_{1}^{k} u\left(x\right)^2 \,dx = 6\pi \) but is unsure how to find \( k \).
  • Another participant proposes a setup involving \( 3\int_{-1}^{1} x+1\,dx=\int_{1}^{k} x+1\,dx \) and suggests using the Fundamental Theorem of Calculus (FTOC) to solve for \( k \).
  • Some participants question whether the problem pertains to volume, confirming that it does relate to solids of revolution and noting that \( \pi \) can be factored out.
  • A participant provides a detailed working through the integration process, leading to the quadratic equation \( k^2 + 2k - 15 = 0 \) and identifies potential solutions for \( k \) as \( 3 \) and \( -5 \), concluding that \( k = 3 \) is the valid solution based on the requirement \( 1 < k \).

Areas of Agreement / Disagreement

Participants generally agree on the method of setting up the integrals and the relevance of the volume concept, but there is some uncertainty regarding the interpretation of the problem and the implications of the solutions for \( k \). Multiple approaches to solving for \( k \) are presented, and while one solution is identified, the discussion remains open to further exploration.

Contextual Notes

Some participants express uncertainty about the volume aspect of the problem and the implications of the integration limits. The discussion does not resolve all assumptions regarding the setup of the integrals or the interpretation of the results.

karush
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$u\left(x\right)=\sqrt{x+1}$

$\pi\int_{-1}^{1} u\left(x\right)^2 \,dx = 2\pi$

$\pi\int_{1}^{k} u\left(x\right)^2 \,dx = 6\pi$

I don't know how to get $k$
 
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The way I would set this up is:

$$3\int_{-1}^{1} x+1\,dx=\int_{1}^{k} x+1\,dx$$

After you apply the FTOC, you should be able to solve for $k$. :)
 
But aren't we dealing with volume?
 
karush said:
But aren't we dealing with volume?

Yes, solids of revolution (disk method), both of which will have $\pi$ as a factor, which we can divide out. :)
 
$u\left(x\right)=\sqrt{x+1}$

so
$\pi\int_{1}^{k} u\left(x\right) \,dx = 6\pi$

$\frac{{k}^{2}}{2}+k-\frac{3}{2}=6$

${k}^{2}+2k-15=0$

so
$k=3\ k=-5$
answer is $k=3$
 
Last edited:
Yes, here's my working:

$$3\int_{-1}^{1} x+1\,dx=\int_{1}^{k} x+1\,dx$$

$$3\left[\frac{x^2}{2}+x\right]_{-1}^1=\left[\frac{x^2}{2}+x\right]_1^k$$

$$3\left(\left(\frac{1}{2}+1\right)-\left(\frac{1}{2}-1\right)\right)=\left(\left(\frac{k^2}{2}+k\right)-\left(\frac{1}{2}+1\right)\right)$$

$$6=\frac{k^2}{2}+k-\frac{3}{2}$$

$$k^2+2k-15=0$$

$$(k-3)(k+5)=0$$

Since we require $1<k$, we conclude that:

$$k=3$$.
 

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