U substitution and partial integration

[James889]

Hey,

I need to evaluate \int_{1}^{5}(6-2x)\sqrt{5-x}dx

So.
\tex{Let ~~u} = 6-2x~~ \tex{then}~~ du = 2

\frac{1}{2}~du = dx
New limits:
x = 5 \longrightarrow u = -4
x = 1 \longrightarrow u = 4

now, -\frac{1}{2}\int^{4}_{-4} u*\sqrt{5-x}

and now for the partial integration.

u \sqrt{5-x}~~ - \int 1\times \sqrt{5-x} = u \sqrt{5-x} ~~-\frac{2}{3}(5-x)^{3/2}

Substituting in my limits of integration:
(6-2*4) \times \sqrt{5-4} ~~-\frac{2}{3}(5-4)^{3/2} - (6-2*(-4)) \times \sqrt{5-(-4)} ~~-\frac{2}{3}(5-(-4))^{3/2} == \tex{wrong!}

what is it that I am doing wrong?
//James

 

[Elucidus]

what is it that I am doing wrong?
//James

Once you've established a relation between u and x, partial integration is not possible (since you are assuming one is constant as the other varies, contradicting the established relation).

Since the original integrand is explicit in x and you're changing into the context of u, you need to change the entire integral into the context of u (nothing in x left behind).

My suggestion is to use the substitution u = 5 - x and go from there. You'll end up with an integrand involving fractional powers of u. The solution should come out to 64/15.

--Elucidus

 

[Mark44]

You have two mistakes in your substitution, one of them being a sign error.

{Let ~~u} = 6-2x~~ \tex{then}~~ du = 2

With this substitution du = - 2dx

Also, I would split the integral into two integrals, like so:
\int_{1}^{5}(6-2x)\sqrt{5-x}dx = \int_1^5 6\sqrt{5 - x}dx ~~-~~ 2\int_1^5 x\sqrt{5 -x}dx
The first integral is straightforward; the second can probably be done with an ordinary substitution.

A final note: what you're calling "partial integration" is actually called "integration by parts," in English. Another integration technique is called "partial fractions decomposition." Similar name, but otherwise very different.

 

[Дьявол]

Actually, you missed something.

If u=6-2x then x=\frac{6-u}{2}

So your integral would be:
<br /> -\frac{1}{2}\int^{4}_{-4} u*\sqrt{5-\frac{6-u}{2}}du<br />

Now use integration by parts.

Regards.

 

[Elucidus]

I see now that you were integrating by parts (as was mentioned in the previous post). Partial integration usually refers to integrating a mutlivaraite function with respect to one of its independent variables a la

\text{If }\frac{\partial}{\partial x}F(x,y) = f(x,y) \text{ then } \int f(x,y) \partial x = F(x, y)

The original attempt letting u = 6 - 2x leads to:

du = -2dx \Rightarrow dx = \frac{du}{-2}

x=1,5 \Rightarrow u=4,-4

=\int_4^{-4} u \sqrt{5-x}\;\frac{du}{-2}

=\frac{1}{2}\int_{-4}^4 u \sqrt{5-x}\;{du}\;\;\text{(Note transpose of boundaries)}

then attempting to apply integration by parts one gets

v = u \Rightarrow dv = du

dw = \sqrt{5-x}\;{du} \Rightarrow w = \text{ ?}

The expression for dw is in terms of x and we're trying to find an anti-derivative with respect to u. This cannot be done without changing the context to fully in x or in u. Mixed contexts lead to errors. I suspect this came about because the differential factor (du) got lost somewhere along the way.

Sorry for the original confusion.

--Elucidus

 

[Dick]

Why don't you try u=5-x instead. That makes x=5-u and it's easy to work out 6-2x in terms of u. It's really much simpler.

 

[James889]

Why don't you try u=5-x instead. That makes x=5-u and it's easy to work out 6-2x in terms of u. It's really much simpler.

That is probably a good idea.

So let u = 5-x ~~\tex{then}~~ du = -1dx
New limits:
x = 5 \longrightarrow u = 0
x = 1 \longrightarrow u = 4
-1\int_{0}^{4}(-6+2u)du~~\times~~u^{1/2}

Does that look reasonable ?

 

[Dick]

That is probably a good idea.

So let u = 5-x ~~\tex{then}~~ du = -1dx
New limits:
x = 5 \longrightarrow u = 0
x = 1 \longrightarrow u = 4
-1\int_{0}^{4}(-6+2u)du~~\times~~u^{1/2}

Does that look reasonable ?

6-2x doesn't turn into (-6+2u) if x=5-u, does it? Check that. And you make me nervous writing the u^(1/2) outside the integral, put it back in and multiply the product out. Then integrate.

 

[James889]

6-2x doesn't turn into (-6+2u) if x=5-u, does it? Check that. And you make me nervous writing the u^(1/2) outside the integral, put it back in and multiply the product out. Then integrate.

Sorry i had a proper brainfart.

(6-2x)~~ \text{in terms of}~~ u~~\text{is}~2u-4

Im also a little confused about the notation. Is the correct notation

2u-4du or just 2u-4 ?

//Jones

 

[Mark44]

Sorry i had a proper brainfart.

(6-2x)~~ \text{in terms of}~~ u~~\text{is}~2u-4

Im also a little confused about the notation. Is the correct notation

2u-4du or just 2u-4 ?

//Jones

2u - 4 is correct. There are actually two sets of equations: one that defines the relationship between dx and du, and one that defines the relationship between x and u.

If u = 5 - x, then du = -dx
u = 5 - x <==> x = 5 - u ==> 6 - 2x = 6 - 2(5 - u) = 2u - 4

 

[Dick]

Sorry i had a proper brainfart.

(6-2x)~~ \text{in terms of}~~ u~~\text{is}~2u-4

Im also a little confused about the notation. Is the correct notation

2u-4du or just 2u-4 ?

//Jones

The correct notation is the integral of (2u-4)*u^(1/2)*du. Expand the product and integrate term by term.

 

[Elucidus]

Quick comment. The limits of integration are reversed. The limits were x = 1 to x = 5. The new "u" values should be u = 4 to u = 0. Limits of integration are not always in ascending order. You can interchange them later by changing the sign of the integral.

--Elucidus

 

[James889]

So.
(2u-4)*u^{1/2} = 2u-4u^{1/2}

Integrating term by term gives u^2-\frac{3}{8}u^{3/2}

~~\left|_{0} ^{4}\left -u^2-\frac{3}{8}u^{3/2}

And the answer is 4.

 

[Elucidus]

So.
(2u-4)*u^{1/2} = 2u-4u^{1/2}

Integrating term by term gives u^2-\frac{3}{8}u^{3/2}

~~\left|_{0} ^{4}\left -u^2-\frac{3}{8}u^{3/2}

And the answer is 4.

Slight mistake:

(2u-4)*u^{1/2} = 2u^{3/2}-4u^{1/2}

The answer, as I mentioned in an earlier post, should be 64/15.

--Elucidus

 

[James889]

Slight mistake:

(2u-4)*u^{1/2} = 2u^{3/2}-4u^{1/2}

The answer, as I mentioned in an earlier post, should be 64/15.

--Elucidus

oops, i always mix up the exponential rules =/