How Can Substitution and Integration Help Solve This Equation?

[cd246]

Homework Statement

/int (2t+4)^-1/2 dt. the answer is 2(sqrt3)-2

Homework Equations

The Attempt at a Solution

u^-1/2*(1/2)du
(-1/2)(u^1/2)
(-1/4)(2t+4)^1/2
(-1/4)(sqrt 12)= (-1/4)(4(sqrt 3)/1)=sqrt 3
(sqrt 3)-1/2
2(sqrt 3)-1

 

[rootX]

Homework Statement

/int (2t+4)^-1/2 dt. the answer is 2(sqrt3)-2

Homework Equations

The Attempt at a Solution

u^-1/2*(1/2)du
(-1/2)(u^1/2)
(-1/4)(2t+4)^1/2
(-1/4)(sqrt 12)= (-1/4)(4(sqrt 3)/1)=sqrt 3
(sqrt 3)-1/2
2(sqrt 3)-1

-You don't need substitution for this.
simply use this exp:

$$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}$$

>also recheck your calculations.

 

[cd246]

I did this: u^(1/2)/(1/2).
2u^(1/2)
2(2t+4)^(1/2)
(4t+8)^1/2
(12t)^(1/2)
2(3)^(1/2)
I believe I am further off than the first time.

 

[CompuChip]

If you don't see how to do the integral with rootX's suggestion, you can turn it around.
What happens if you differentiate $$x^{n + 1}/(n + 1)$$ with $x = (2t + 4)$ and $n = -1/2$?
Can you find a primitive now? (Watch the chain rule, you might need to add some constants to get the integrand back correctly)

 

[malawi_glenn]

What are the integration limits? From what to what are the integration performed over?

 

[HallsofIvy]

Homework Statement

/int (2t+4)^-1/2 dt. the answer is 2(sqrt3)-2

Well, no, that is not the answer! The answer is (2t+4)^1/2+ C where C can be any constant. Is it possible that your integral had limits that you haven't told us?

Homework Equations

The Attempt at a Solution

u^-1/2*(1/2)du

Please, please, pleas, tell us what you are doing! This makes no sense until you tell us that you are using the substitution u= 2x- 4 so that (2x-4)^-1/2 becomes u^-1/2 (not ^1/2) and 2dx= du so dx= (1/2)du

(-1/2)(u^1/2)

?? The anti-derivative of uⁿ is 1/(n+1) uⁿ⁺¹ (as long as n is not -1). For n= -1/2, n+1= 1/2 and that becomes 2 u^1/2, Multiplying by 1/2 from the "dx= (1/2) du" gives u^1/2+ C= (2t+4)^1/2+ C.

(-1/4)(2t+4)^1/2
(-1/4)(sqrt 12)= (-1/4)(4(sqrt 3)/1)=sqrt 3
(sqrt 3)-1/2
2(sqrt 3)-1

 

[cd246]

Well, no, that is not the answer! The answer is (2t+4)^1/2+ C where C can be any constant. Is it possible that your integral had limits that you haven't told us?
Please, please, pleas, tell us what you are doing! This makes no sense until you tell us that you are using the substitution u= 2x- 4 so that (2x-4)^-1/2 becomes u^-1/2 (not ^1/2) and 2dx= du so dx= (1/2)du

?? The anti-derivative of uⁿ is 1/(n+1) uⁿ⁺¹ (as long as n is not -1). For n= -1/2, n+1= 1/2 and that becomes 2 u^1/2, Multiplying by 1/2 from the "dx= (1/2) du" gives u^1/2+ C= (2t+4)^1/2+ C.

I cannot believe this, same problem but it is /int 4_0. I'm sorry i missed that.

To answer the question of what i am doing, first I have to figure out if this is something i can use the fund. rules of integrals or something i have to (or is preferred) substitute the function. For this one, i believe that i have to use sub.

 

[malawi_glenn]

And yes people here have showing you that make the subsitute:

u = 2t + 4

Makes the problem slightly easier, have you tried this? And You do not seem to understand the basic algebraic rules here:

2(2t+4)^(1/2)
(4t+8)^1/2

in your post #3

Now this is true:
$$a^{1/2} * b^{1/2} = (ab)^{1/2}$$
So:
$$2(2t+4)^{1/2} = (4^{1/2})(2t+4)^{1/2} =(8t+16)^{1/2}$$

Anyway, if you make that substitution u, how does you new integral looks like?

 

[HallsofIvy]

I think rootX's point:

-You don't need substitution for this.
simply use this exp:

$$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}$$

>also recheck your calculations.

is that for simple linear substitutions, ax+ b, you should be able to just think: do the integral ignoring the linear part inside the power, and the divide by a: The anti-derivative of (ax+ b)ⁿ is (n/a)(ax+ b)_{n+1[/sup]. You don't really need to write out the substitution. That comes with experience.}

 

[cd246]

i finally found my error. I understand the basics(for the most part) of integration.
It is the substitution and the more sophisticated equations that get me.