1. intergration, i m stuck.

can someone help me intergrate this step by step

2. Originally Posted by Legendsn3verdie
can someone help me intergrate this step by step
Two hints (sorry but I don't do step-by-step as a general rule):

1. The integral can be written $\displaystyle \frac{1}{\sqrt{2}} \int_{1}^{3} (-x + 3)^{1/2} \, dx$.

2. You should have been taught that $\displaystyle \int (ax + b)^n \, dx = \frac{1}{a} \, \frac{1}{(n+1)} (ax + b)^{n+1} + C$ provided $\displaystyle n \neq -1$.

3. Originally Posted by mr fantastic
Two hints (sorry but I don't do step-by-step as a general rule):

1. The integral can be written $\displaystyle \frac{1}{\sqrt{2}} \int_{1}^{3} (-x + 3)^{1/2} \, dx$.

2. You should have been taught that $\displaystyle \int (ax + b)^n \, dx = \frac{1}{a} \, \frac{1}{(n+1)} (ax + b)^{n+1} + C$ provided $\displaystyle n \neq -1$.

hmm dunno that proof u mentioned .. is there a way to do this with u subsitituion?> i keep seeming to juggle a 1/2 around.

u = 3-x
du = -dx/2?

and just put a 2 outside of the integral so they cancel out.. is that a valid way?

4. Originally Posted by Legendsn3verdie
hmm dunno that proof u mentioned .. is there a way to do this with u subsitituion?> i keep seeming to juggle a 1/2 around.

[snip]
u = 3-x
du = -dx/2?

and just put a 2 outside of the integral so they cancel out.. is that a valid way?
Let $\displaystyle u = \frac{3 - x}{2}$. Then $\displaystyle \frac{du}{dx} = - \frac{1}{2} \Rightarrow dx = -2 du$. Note also that $\displaystyle x = 1 \Rightarrow u = 1$ and $\displaystyle x = 3 \Rightarrow u = 0$. Substitute all this in and solve the new integral.

5. Originally Posted by mr fantastic
Let $\displaystyle u = \frac{3 - x}{2}$. Then $\displaystyle \frac{du}{dx} = - \frac{1}{2} \Rightarrow dx = -2 du$. Note also that $\displaystyle x = 1 \Rightarrow u = 1$ and $\displaystyle x = 3 \Rightarrow u = 0$. Substitute all this in and solve the new integral.

ah i see i got 2 for a answer with that sub. and it is right, thanks, you are brillant.

my only question to that is.. how would u know to sub U in for the whole (3-x)/2.. to me that kinda seems strange cause once u sub u for that, there is nothing really left!, and when u take there derivative of U to get du, u dont see a -1/2 or anything anywhere to relate it to.. so i guess my question summ'd up is, how do u know to sub U as 3-x/2?

6. is it because their is nothing else i can do? i remember my old teacher always saying that.

7. Originally Posted by Legendsn3verdie
ah i see i got 2 for a answer with that sub. and it is right, thanks, you are brillant.

my only question to that is.. how would u know to sub U in for the whole (3-x)/2.. to me that kinda seems strange cause once u sub u for that, there is nothing really left!, and when u take there derivative of U to get du, u dont see a -1/2 or anything anywhere to relate it to.. so i guess my question summ'd up is, how do u know to sub U as 3-x/2?
The technique is called linear substitution. I knew to use it because $\displaystyle \frac{3-x}{2}$ is a linear function of x.

8. Originally Posted by mr fantastic
The technique is called linear substitution. I knew to use it because $\displaystyle \frac{3-x}{2}$ is a linear function of x.
but still what about 3-x over 2 makes it a linear function.. how would someonelike me (who isnt so great at math) know to just make u = 3 -x /2.. cause if i didnt come to this forum i d be tryin intgration by parts, u = 3- x, partial fractions all kinds of wastes of time..

9. Originally Posted by Legendsn3verdie
but still what about 3-x over 2 makes it a linear function.. how would someonelike me (who isnt so great at math) know to just make u = 3 -x /2.. cause if i didnt come to this forum i d be tryin intgration by parts, u = 3- x, partial fractions all kinds of wastes of time..
Experience.