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- Homework Statement
- See attached textbook question and solution

- Relevant Equations
- Integration - Calculus

This is the question...hmmmm it stressed me a little bit.

Find the textbook solution here;

Now my approach to this was as follows;

On integration,

##y=\dfrac{(kx-1)^6}{3k} +c##

on using the point ##(0,1)## and ##(1,8)##, we end up with

##1=\dfrac{1}{3k} +c##

##8=\dfrac{(k-1)^6}{3k} +c##

it follows that,

##8=\dfrac{(k-1)^6}{3k} + \dfrac{3k-1}{3k} ##

...

##(k-1)^6+3k-1-24k=0##

##(k-1)^6-21k-1=0##

##k^6-6k^5+15k^4-20k^3+15k^2-6k+1-21k-1=0##

I let ##P(k)=k^6-6k^5+15k^4-20k^3+15k^2-6k+1-21k-1##

using trial and error method from ##±0, ±1...## i got ##P(3)=0## implying ##k=3## is a factor,

thus substituting on;

##1=\dfrac{1}{3k} +c##

we get;

##1=\dfrac{1}{3×3} +c##

##1=\dfrac{1}{9} +c##

##c=\dfrac{8}{9}##

Therefore;

##y=\dfrac{(3x-1)^6}{3×3} +\dfrac{8}{9}##

##y=\dfrac{(3x-1)^6+8}{9}##

I would be interested if there would be another approach to this question. Cheers guys.

Find the textbook solution here;

**no. 6**Now my approach to this was as follows;

On integration,

##y=\dfrac{(kx-1)^6}{3k} +c##

on using the point ##(0,1)## and ##(1,8)##, we end up with

##1=\dfrac{1}{3k} +c##

##8=\dfrac{(k-1)^6}{3k} +c##

it follows that,

##8=\dfrac{(k-1)^6}{3k} + \dfrac{3k-1}{3k} ##

...

##(k-1)^6+3k-1-24k=0##

##(k-1)^6-21k-1=0##

##k^6-6k^5+15k^4-20k^3+15k^2-6k+1-21k-1=0##

I let ##P(k)=k^6-6k^5+15k^4-20k^3+15k^2-6k+1-21k-1##

using trial and error method from ##±0, ±1...## i got ##P(3)=0## implying ##k=3## is a factor,

thus substituting on;

##1=\dfrac{1}{3k} +c##

we get;

##1=\dfrac{1}{3×3} +c##

##1=\dfrac{1}{9} +c##

##c=\dfrac{8}{9}##

Therefore;

##y=\dfrac{(3x-1)^6}{3×3} +\dfrac{8}{9}##

##y=\dfrac{(3x-1)^6+8}{9}##

I would be interested if there would be another approach to this question. Cheers guys.

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