Question

Need some help for a friend (He doesn't have access to a computer)

Question in reply...

Answer 1

\(\large\sf\int{(4t+3)^3}dt\ =\ \color{blue}{\dfrac{1}4}*\color{red}{\dfrac{1}4}(4t+3)^\color{red}4+C\)

Can anyone explain the blue part? Why is it necessary to put an extra 1/4 in front. (probably has something to do with 4t I guess)

Answer 2

It may be easiest to see why it has to be there, if you take the derivative of the result:

\[\Large \frac{ dy }{dx } \ \color{blue}{\dfrac{1}4}*\color{red}{\dfrac{1}4}(4t+3)^\color{red}4= \frac{ 1 }{16 }\cdot 4\left( 4t+3 \right)^3\cdot4\]
\[\Large =16\cdot \frac{ 1 }{16 }\left( 4t+3 \right)^3=\left( 4t+3 \right)^3\]

Answer 3

sorry, that should be "dy/dt" of course. :)

Answer 4

Yea, but when I had to answer \(\large\sf\int{(4t+3)^3}dt\), how am i supposed to know that I have to add an extra 1/4?
Because there's a 4 in front of t?

Answer 5

Yes, because of the 4 in front of the t.

It's a nice thing to try and get used to with integration.
Example:\[\Large \int\limits \cos(2x)dx \qquad=\qquad \frac{1}{2}\sin(2x)\]
Normally, the derivative would produce an extra factor of 2, because of the 2x right?
Well the opposite happens when we integrate, we have to divide by the factor of 2.

Answer 6

You would do it by substitution.... let

u=4t+3
Then du=4dt

But that gives me:

\(\large\sf\int{(4t+3)^3}dt\ =\sf\int{u^3}du=\sf\int{(4t+3)^3}\color{red}4dt\)

OOPS I have an extra factor of 4.... but if I rewrite the original integral:

\(\large\sf\int{\dfrac{4}{4} (4t+3)^3}dt\)
which is of course equivalent to the original, then I have:
\(\large\sf\int{\dfrac{4}{4} (4t+3)^3}dt=\dfrac{1}{4}\sf\int{4\cdot (4t+3)^3}dt\)

And now my substitution above works perfectly:
\(\dfrac{1}{4}\sf\int{4\cdot (4t+3)^3}dt=\dfrac{1}{4}\sf\int{u^3 du}\)

Neato, huh? :)

Answer 7

Alright,
So:
\(\Large\int(x+5)^4dx\ =\ \dfrac{1}5(x+5)^5+C\)
And:
\(\Large\int(7x+5)^4dx\ =\ \dfrac{1}7*\dfrac{1}5(7x+5)^5+C\)

Answer 8

ya looks good :)

Answer 9

Oke, Thanks @DebbieG and @zepdrix \(\LARGE\ddot\smile\)

Answer 10

Sorry, was typing and then had a phone call... but I can't leave this unfinished, lol...

So with the substitution we get your result above:

\(\Large \dfrac{1}{4}\sf\int{u^3 du}=\dfrac{1}{4}\cdot \dfrac{1}{4}u^4 =\dfrac{1}{4}\cdot \dfrac{1}{4}(4t+3)^4 +C\)

although certainly, a lot of these are easy enough to ignore the extra factor and do the integral, then figure out what factor you need to use to "make it work". :)