# U substitution and partial integration

1. Aug 10, 2009

### 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 im doing wrong?
//James

2. Aug 10, 2009

### Elucidus

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

3. Aug 10, 2009

### Staff: Mentor

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

4. Aug 10, 2009

### Дьявол

Actually, you missed something.

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

$$-\frac{1}{2}\int^{4}_{-4} u*\sqrt{5-\frac{6-u}{2}}du$$

Now use integration by parts.

Regards.

5. Aug 10, 2009

### 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

6. Aug 10, 2009

### 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.

7. Aug 11, 2009

### James889

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 ?

Last edited: Aug 11, 2009
8. Aug 11, 2009

### Dick

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.

9. Aug 11, 2009

### James889

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

10. Aug 11, 2009

### Staff: Mentor

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

11. Aug 11, 2009

### Dick

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

12. Aug 11, 2009

### 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

13. Aug 12, 2009

### James889

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

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

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

Last edited: Aug 12, 2009
14. Aug 12, 2009

### Elucidus

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

15. Aug 12, 2009

### James889

oops, i always mix up the exponential rules =/