Definite Integral Example - # 2

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Jason76

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\(\displaystyle \int_{1}^{9} \dfrac{x}{\sqrt{1 + 2x}} dx\)

\(\displaystyle \int_{1}^{9} \dfrac{x}{(1 + x)^{1/2}}dx\)

\(\displaystyle \int_{1}^{9} x(1 + x)^{-1/2} dx\)

\(\displaystyle \int_{1}^{9} x(u)^{-1/2} dx\)


\(\displaystyle u = 1 + 2x\)

\(\displaystyle du = 2 dx\) - cannot use fraction multiplication to manipulate this.

So

\(\displaystyle u = 1 + 2x\)

\(\displaystyle u - 1 = 2x\) - solve for x

\(\displaystyle \dfrac{u}{2} - \dfrac{1}{2} = x\) How to finish algebra on this?
 
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Jason76 said:
\(\displaystyle \int_{1}^{9} \dfrac{x}{\sqrt{1 + 2x}} dx\)

\(\displaystyle \int_{1}^{9} \dfrac{x}{(1 + x)^{1/2}}dx\)
Click to expand...
What happened to the "2" inside the radical?

Jason76 said:
\(\displaystyle u = 1 + 2x\)

\(\displaystyle du = 2 dx\) - cannot use fraction multiplication to manipulate this.
Click to expand...
What does that statement mean? Are you saying that you don't know how to divide through, or that the instructions (which you have declined to include) direct you to do otherwise?

Jason76 said:
\(\displaystyle \dfrac{u}{2} - \dfrac{1}{2} = x\) How to finish algebra on this?
Click to expand...
Dunno. What is the "this" that you're doing the algebra to? What is the point? What are you trying to accomplish?

Seriously, we really can't read your mind. ;)
 
stapel said:
What happened to the "2" inside the radical?


What does that statement mean? Are you saying that you don't know how to divide through, or that the instructions (which you have declined to include) direct you to do otherwise?


Dunno. What is the "this" that you're doing the algebra to? What is the point? What are you trying to accomplish?

Seriously, we really can't read your mind. ;)
Click to expand...

goal is to isolate x (which is already done), and provide something meaningful on the other side.
 
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\(\displaystyle \int_{1}^{9} \dfrac{x}{\sqrt{1 + 2x}} dx\)

\(\displaystyle \int_{1}^{9} \dfrac{x}{(1 + x)^{1/2}}dx\)

\(\displaystyle \int_{1}^{9} x(1 + x)^{-1/2} dx\)

\(\displaystyle \int_{1}^{9} x(u)^{-1/2} dx\)


\(\displaystyle u = 1 + 2x\)

\(\displaystyle du = 2 dx\) - cannot use fraction multiplication to manipulate this.

So

making new limits of integration: Adding this part......

\(\displaystyle u = 1 + 2(1) = 3\)

\(\displaystyle u = 1 + 2(9) = 19\)

\(\displaystyle u = 1 + 2x\)

\(\displaystyle u - 1 = 2x\) - solve for x

\(\displaystyle \dfrac{u}{2} - \dfrac{1}{2} = x\) How to finish algebra on this?
 
Then again you, you don't have to make new limits of integration (it's optional) on any integration problem, but you do have to solve (the u equation) for x, in the case of these special ones. Is this correct?
 
Jason76 said:
Then again you, you don't have to make new limits of integration (it's optional) on any integration problem, but you do have to solve (the u equation) for x, in the case of these special ones. Is this correct?
Click to expand...
Um... what?
 
Jason76 said:
Then again you, you don't have to make new limits of integration (it's optional) on any integration problem, but you do have to solve (the u equation) for x, in the case of these special ones. Is this correct?
Click to expand...
If, with a definite integral, you change from variable "x" to variable "u", you can either change the limits from their "x" value to the corresponding "u" value and evaluate or change back to "x" and use the original limits. Most often, the first is easier.

For example, to integrate \(\displaystyle \int_0^1 (x+ 3)^2 dx\), we can let u= x+3, du= dx, to get \(\displaystyle \int u^2 du\).

Now the simpler thing to do would be to change the limits of integration: when x= 0, u=0+ 3= 3 and when x= 1, u= 1+ 3= 4 so the integral becomes \(\displaystyle \int_0^1 (x+ 3)^2 dx= \int_3^4 u^2 du= \left[\frac{1}{3} u^3\right]_3^4= \frac{1}{3}(64- 27)= \frac{37}{3}\).

But we can also just do the "indefinite integral", \(\displaystyle \int u^2 du= \frac{1}{3} u^3\), and then replace u with x+ 3 again: \(\displaystyle \left[\frac{1}{3}(x+ 3)^3\right]_0^1= \frac{1}{3}((1+ 3)^3- (0+ 3)^3)= \frac{1}{3}(64- 27)= \frac{37}{3}\) just as before.

(Of course, in this simple case we could just multiply: \(\displaystyle (x+ 3)^2= x^2+ 6x+ 9\) so that \(\displaystyle \int_0^1 (x+ 3)^2 dx= \int_0^1 x^2+ 6x+ 9 dx= \left[\frac{1}{3}(x^3+ 3x^2+ 9x)\right]_0^1= \frac{1}{3}+ 3+ 9= \frac{1}{3}+ \frac{9}{3}+ \frac{27}{3}= \frac{37}{3}\).)
 
\(\displaystyle u = 1 + 2x\)

\(\displaystyle u - 1 = 2x\) - solve for x
Click to expand...

Regardless of whether limits of integration are being changed, the above step is needed when fraction multiplication can't be used, or the dx is an exact match of what is in the original integral. :confused:
 
Jason76 said:
Regardless of whether limits of integration are being changed, the above step is needed when fraction multiplication can't be used, or the dx is an exact match of what is in the original integral.
Click to expand...
What do you mean by "when fraction multiplication can't be used"? Is the above statement meant to be a question of some sort? What do you want?

Seriously, we really can't read your mind. ;)
 
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