Integration by recognition, explanation needed

Sep 2008
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2
I am trying to understand the example in my book, however can't seem to makc sense of it. If someone could just explain it to me.

\(\displaystyle \frac{d}{dx}(3x-2)^{5} = 15(3x-2)^{4} \)

\(\displaystyle \int 15(3x-2)^{4} dx = (3x-2)^{5} + C \)

the next line I don't get, why do they divide both sides by 15? What's the need?

the book says 'an adjustment factor of 1/15 is needed'.

\(\displaystyle \int (3x-2)^{4} dx = \frac{1}{15}(3x-2)^{5} + c \)
 
Dec 2009
3,120
1,342
I am trying to understand the example in my book, however can't seem to makc sense of it. If someone could just explain it to me.

\(\displaystyle \frac{d}{dx}(3x-2)^{5} = 15(3x-2)^{4} \)

\(\displaystyle \int 15(3x-2)^{4} dx = (3x-2)^{5} + C \)

the next line I don't get, why do they divide both sides by 15? What's the need?

the book says 'an adjustment factor of 1/15 is needed'.

\(\displaystyle \int (3x-2)^{4} dx = \frac{1}{15}(3x-2)^{5} + c \)
If evaluating the above line only, Tweety...

\(\displaystyle \int{(3x-2)^4}dx=\frac{1}{15}(3x-2)^5+C\)

because of \(\displaystyle \frac{d}{dx}\left[(3x-2)^5+C\right]=15(3x-2)^4\)

hence \(\displaystyle \frac{1}{15}\ \frac{d}{dx}\left[(3x-2)^5+C\right]=(3x-2)^4\)

therefore \(\displaystyle \int{(3x-2)^4}dx=\frac{1}{15}\left[(3x-2)^5+C\right]\)

You can try it using substitution of course... \(\displaystyle u=3x-2,\ du=3dx,\ \frac{du}{3}=dx\)

\(\displaystyle \int{u^4}\frac{du}{3}=\frac{1}{3}\int{u^4}du=\frac{1}{3}\ \frac{1}{5}\ u^5+c=\frac{1}{15}(3x-2)^5+C\)
 
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Sep 2008
631
2
If evaluating the above line only, Tweety...

\(\displaystyle \int{(3x-2)^4}dx=\frac{1}{15}(3x-2)^5+C\)

because of \(\displaystyle \frac{d}{dx}\left[(3x-2)^5+C\right]=15(3x-2)^4\)

hence \(\displaystyle \frac{1}{15}\ \frac{d}{dx}\left[(3x-2)^5+C\right]=(3x-2)^4\)

therefore \(\displaystyle \int{(3x-2)^4}dx=\frac{1}{15}\left[(3x-2)^5+C\right]\)

You can try it using substitution of course... \(\displaystyle u=3x-2,\ du=3dx,\ \frac{du}{3}=dx\)

\(\displaystyle \int{u^4}\frac{du}{3}=\frac{1}{3}\int{u^4}du=\frac{1}{3}\ \frac{1}{5}\ u^5+c=\frac{1}{15}(3x-2)^5+C\)

Thank you,

However I am still unclear about this, because we are trying to work out the \(\displaystyle \int 15(3x-2)^{4} dx \)

not \(\displaystyle \int (3x-2)^{4} dx \)

for example, \(\displaystyle \frac{d}{dx} sin3x = cos3x \times 3 \)

so

\(\displaystyle \int cos3x \times 3dx = sin3x + c \)

so shouldn't it just be \(\displaystyle \int 15(3x-2)^{4} dx = (3x-2)^{5} + C \) ?
 
Dec 2009
3,120
1,342
Thank you,

However I am still unclear about this, because we are trying to work out the \(\displaystyle \int 15(3x-2)^{4} dx \)

not \(\displaystyle \int (3x-2)^{4} dx \)

for example, \(\displaystyle \frac{d}{dx} sin3x = cos3x \times 3 \)

so

\(\displaystyle \int cos3x \times 3dx = sin3x + c \)

so shouldn't it just be \(\displaystyle \int 15(3x-2)^{4} dx = (3x-2)^{5} + C \) ?
Hi Tweety,

I think the book example is calculating \(\displaystyle \int{(3x-2)^4}dx\) using the result from \(\displaystyle \int{15(3x-2)^4}dx\)

since \(\displaystyle \int{15(3x-2)^4}dx=15\int{(3x-2)^4}dx\)
 
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