Integrate this fraction please :)

Integrate 1/sqrt(3x-2) between bounds 2 and 1

I started to use u=3x-2
du/dx = 3
thus 1/3 du = dx

I then adjusted bounds using 3x-2, to get 6-2=4 and 3-2=1

In terms of u this gave me,

2/3 *integral with bounds 4 and 1* 1/sqrt(u) du

this gave me 1/3(u^0.5) *bounds 4 and 1*
this gave me 1/3 x 1 = 1/3.

Where have I gone wrong? Also apologies about the workings, I couldnt found an integral sign ect...

Reply 1

nerak99

13

You need to divide by the new power

\frac{1}{2}

Reply 2

Slowbro93

20

Original post by jumblehunter

Integrate 1/sqrt(3x-2) between bounds 2 and 1

I started to use u=3x-2
du/dx = 3
thus 1/3 du = dx

I then adjusted bounds using 3x-2, to get 6-2=4 and 3-2=1

In terms of u this gave me,

2/3 *integral with bounds 4 and 1* 1/sqrt(u) du

this gave me 1/3(u^0.5) *bounds 4 and 1*
this gave me 1/3 x 1 = 1/3.

Where have I gone wrong? Also apologies about the workings, I couldnt found an integral sign ect...

You've forgotten to divide your answer by your new power :yep:

Reply 3

Star-girl

12

Original post by jumblehunter

Integrate 1/sqrt(3x-2) between bounds 2 and 1

I started to use u=3x-2
du/dx = 3
thus 1/3 du = dx

I then adjusted bounds using 3x-2, to get 6-2=4 and 3-2=1

In terms of u this gave me,

2/3 *integral with bounds 4 and 1* 1/sqrt(u) du

this gave me 1/3(u^0.5) *bounds 4 and 1*
this gave me 1/3 x 1 = 1/3.

Where have I gone wrong? Also apologies about the workings, I couldnt found an integral sign ect...

The others have told you were it went wrong but you might like to know that if you rewrite the integrand like this:

\displaystyle \int^2_1 \frac{1}{\sqrt{3x-2} } \ dx = \int^2_1 (3x-2)^{-\frac{1}{2} } \ dx

Then it is clear that since the derivative of the expression in brackets is a constant, you can integrate this directly to

\displaystyle \left[\frac{2}{3} (3x-2)^{\frac{1}{2} } \right]^2_1 = \frac{2}{3} \left[(3x-2)^{\frac{1}{2} } \right]^2_1

...and continue from there thus making it easier for yourself instead of using a substitution (although both methods are valid of course).

(edited 11 years ago)

Reply 4

jumblehunter

OP

8

Original post by Star-girl

The others have told you were it went wrong but you might like to know that if you rewrite the integrand like this:

\displaystyle \int^2_1 \frac{1}{\sqrt{3x-2} } \ dx = \int^2_1 (3x-2)^{-\frac{1}{2} } \ dx

Then it is clear that since the derivative of the expression in brackets is a constant, you can integrate this directly to

\displaystyle \left[\frac{2}{3} (3x-2)^{\frac{1}{2} } \right]^2_1 = \frac{2}{3} \left[(3x-2)^{\frac{1}{2} } \right]^2_1

...and continue from there thus making it easier for yourself instead of using a substitution (although both methods are valid of course).

Cheers, but the question specified to use substitution. Would you be able to rewrite that without the shortcuts please?

Reply 5

Star-girl

12

Original post by jumblehunter

Cheers, but the question specified to use substitution. Would you be able to rewrite that without the shortcuts please?

Ah, OK - fair enough then. :smile:

I didn't use a shortcut but I shall explain the steps if you like. Look at the expression

1(3x-2)^{-\frac{1}{2} }

(I've written the 1 in for a reason). When you differentiate

3x-2

, you get

3

. Note that

1 = 3 \times \dfrac{1}{3}

. Since the expression is of the form:

multiple \ of \ derivative \ of \ function \times (function)^{some \ power}

, it means that the chain rule must have been used to get the expression you are trying to integrate, so you can just raise the power on the bracket and then divide by the new power. However, since you had a multiple of the derivative of the function outside the bracket rather than the derivative itself, you have to correct for this. You do it by multiplying what you have by the number that you would need to multiply the derivative by to get the number you had outside the bracket, i.e.

\dfrac{1}{3}

in this case:

\displaystyle \int^2_1 (3x-2)^{-\frac{1}{2} } \ dx = \left[ \frac{(3x-2)^{\frac{1}{2} } }{\frac{1}{2} } \times \frac{1}{3} \right]^2_1 = \left[\frac{2}{3} (3x-2)^{\frac{1}{2} } \right]^2_1

and since when you have a constant inside the square brackets you can take it outside, then the above is the same as

\dfrac{2}{3} \left[(3x-2)^{\frac{1}{2} } \right]^2_1

.

This technique of recognising that the integrand (the function to be integrated) is a composite function X a multiple of its derivative and so integrating directly, is called integration by recognition.

(edited 11 years ago)

Reply 6

jumblehunter

OP

8

"When you differentiate 3x-2, you get 3 and 1 = 3 \times \dfrac{1}{3} . Since the expression is of the form:"

Pardon me if I'm being a muppet, but surely differentiation 3x -2 just gives you 3 -0. I though differentiating integers gives you zero.

Reply 7

Star-girl

12

Original post by jumblehunter

"When you differentiate 3x-2, you get 3 and 1 = 3 \times \dfrac{1}{3} . Since the expression is of the form:"

Pardon me if I'm being a muppet, but surely differentiation 3x -2 just gives you 3 -0. I though differentiating integers gives you zero.

... 3-0 = 3.

Integrate this fraction please :)

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