Question

Can someone help me with Cal 2? int. by parts

Answer 1

you got a specific question?

Answer 2

I have two examples

Answer 3

general idea is
\[\int udv=uv-\int vdu\]

Answer 4

\[\int\limits x(x+7)^{30}dx\]

Answer 5

i think ( i suck at these) that you do not use parts for this, but rather a u - sub lets try that

Answer 6

put \(u=x+7, du = dx, x=u-7\) and turn it in to
\[\int (u-7)u^{30}dyu=\int u^{31}du -\int 7u^{30}du\]

Answer 7

typo there \[\int (u-7)u^{30}du=\int u^{31}du -\int 7u^{30}du\]

Answer 8

my u=x and my v1= (x+7)30

Answer 9

trying to figure out my u1= and v=

Answer 10

i really don't think that will make it any easier
use the u -sub and you should be fine

Answer 11

lets try it
\[\int udv=uv-\int vdu\]
\[u=x, du = dx, dv =(x+7)^{30}, v = \frac{1}{31}(x+7)^{31}\]

Answer 12

then you get
\[\frac{1}{31}x(x+7)^{31}-\frac{1}{31}\int (x+7)^{31}dx\]

Answer 13

that might actually work, i see what you would get the same answer as the one i would get above

Answer 14

where does the 31 come from??

Answer 15

You have to increase the exponent by 1 and divide it by th same number,
so 30+1 = 31.

Answer 16

Okay so can you show me a step by step to get the final answer? I am a little lost.

Answer 17

Satellites steps were pretty clear, were you confused by them?
I'll go into a little bit more detail if you need.

\[\large \int\limits \color{green}{x}(\color{royalblue}{x+7})^{30}\color{purple}{dx}\]

Make a u-substitution,\[\large \color{royalblue}{u=x+7}\]Within our substitution here, let's subtract 7 from each side,\[\large \color{green}{x=u-7}\]
We also want to take the derivative of the blue term,\[\large \color{purple}{du=dx}\]

Plugging all the pieces in gives us,\[\large \int\limits\limits \color{green}{x}(\color{royalblue}{x+7})^{30}\color{purple}{dx} \qquad=\qquad \large \int\limits\limits \color{green}{(u-7)}(\color{royalblue}{u})^{30}\color{purple}{du}\]
Make sense so far?
This method is U-substitution.
If you prefer Integration By Parts, let me know so I don't waste my time with this method XD

Answer 18

ok it starting to make more sense with the different colors

Answer 19

From here, we distribute the u^30 to each term in the first set of brackets.\[\large \int\limits u\cdot u^{30}-7u^{30}\;du\]So this is what we're left integrating.\[\large \int\limits u^{31}-7u^{30}\;du\]

From here we just apply the `Power Rule for Integration`.
Remember how to do that?

Answer 20

now i see how you get 31

Answer 21

you add the U's

Answer 22

Yah it's a rule of exponents that is good to remember! :)\[\large x^2\cdot x^3=x^{2+3}\]

Answer 23

Okay

Answer 24

our next step would be?

Answer 25

Apply the Power Rule for Integration to each term.

Example:\[\large \int\limits x^2\;dx \qquad=\qquad \frac{x^{2+1}}{2+1} \qquad=\qquad \frac{1}{3}x^3\]

2 Steps:
~Increase exponent by 1.
~Divide by this new exponent.

Answer 26

@zepdrix the boss!

Answer 27

:O

Answer 28

O:

Answer 29

Simmer down Genghis XD

Answer 30

ok :)

Answer 31

Did you try it? :D What'd you get?

Answer 32

1/31

Answer 33

:(

Answer 34

wut..?

Answer 35

1/31?
You should get more than just that :o
Where are you u's and stuff?

Answer 36

Lol. Nothing.

Answer 37

\[(u ^{30}-7^{30}) (u)^{30}\]

Answer 38

and 1 for the u outside the bracket \[u(u ^{30}-7^{30})(u)^{30}\]

Answer 39

I can't figure out what you're doing... hmm

Answer 40

If we look at just the first term we had,\[\large \int\limits u^{31}du\]We increase the exponent by 1, then divide by this new exponent.\[\large =\qquad\frac{u^{31+1}}{31+1} \qquad=\qquad\frac{1}{32}u^{32}\]

Answer 41

okay then what is next

Answer 42

Try to do the same for the next term.\[\large \int\limits 7u^{30}du\]The 7 is a constant, let's pull it out of the integral and ignore it for now.\[\large 7\int\limits u^{30}du\]So applying the rule above, what do you get for this one? :o

Answer 43

\[\frac{ 1 }{ 31 }u ^{31}\]

Answer 44

k good.

So our integral gives us,\[\large \int\limits u^{31}-7u^{30}\;du \qquad=\qquad \frac{1}{32}u^{32}-\frac{7}{31}u^{31}+C\]

Answer 45

Then as a final step, undo your substitution.
Replace your u's with (x+7)'s.

Answer 46

so to get 32 you do that step over again?

Answer 47

yes, apply that rule to the first term gave us the 32 power and fraction.

Answer 48

i have one more problem can you help me?