Problems with u-sub core 4.

I am having problems doing problems when I am given a u as in a form u^2 = this and that.

If I were to pick my own value, I'd usually just pick "u", even if it's rooted. But whenever I have to take it squared, or u rooted, I encounter problems.

Usually I just integrate, and then substitute for dx.

With u^2's I get really confused.

For example, in c4, exercise 6H number 4.

I am given x√(2-x) dx

There needs to be an EXACT value to be found, and all that has to be integrated with limits from 2 to 0.

While I did took u as a 2-x, I could only find the approximate value, while when looking at the worked answer, it was suggested I take u^2. I would have never though of that.. and I am struggling in knowing when to take u^2 or rooted u or normal u.

Because by taking u^2, you'd end up with a surd fraction, and it will also make you change limits with one of them being root 2.

So yeah.. I am really having troubles with these... any advice from someone who is good or has experienced the same thing?

Reply 1

Smaug123

Original post by studentFlipper

What was wrong with

u = \sqrt{2-x}

? It's the same substitution as

u^2 = 2-x

, and you'll go through the same motions to evaluate the integral. Would you mind posting your working where you try using the

u

rather than

u^2

substitution?

Reply 2

TenOfThem

Original post by studentFlipper

I just did this with u=2-x
Why would you get an approximate answer?

The limits change from 0,2 to 2,0 and you have some square roots of u
So the answer is a multiple of root2

Reply 3

Smaug123

Original post by studentFlipper

<snip>

(To link between my answer and TenOfThem's, usually more than one substitution will work.)

Reply 4

studentFlipper

Original post by Smaug123

What was wrong with

u = \sqrt{2-x}

? It's the same substitution as

u^2 = 2-x

, and you'll go through the same motions to evaluate the integral. Would you mind posting your working where you try using the

u

rather than

u^2

substitution?

Original post by TenOfThem

I just did this with u=2-x
Why would you get an approximate answer?

The limits change from 0,2 to 2,0 and you have some square roots of u
So the answer is a multiple of root2

Well for me it's not that easy taking u^2, it does confuse me. I am really struggling with math, and sometimes I spent hours staring at a problem or even at a worked out example without really understanding.

Anyway, I drew the worked example of taking 2-x as a u, but I probably messed up with signs on the way or broke some rules.

OnPaste.20150113-073636.png36.1KB

Reply 5

davros

Original post by studentFlipper

I haven't checked your working, but why have you converted everything to decimals?

You should know that

2^{3/2} = 2 \times 2^{1/2} = 2\sqrt{2}

and similarly

2^{5/2} = 2^2 \times 2^{1/2} = 4\sqrt{2}

so you can write everything as a multiple of

\sqrt{2}

(also your notation is slightly wrong - you should have written dx = -du not dx = -1 so that your 'u integral' ends with -du instead of -1. Always make it clear what you're integrating with respect to)

Reply 6

TenOfThem

Original post by studentFlipper

As I said you have everything in root2s

Do not change to decimal

Reply 7

Smaug123

Original post by studentFlipper

So the

u^2 = 2-x

substitution is essentially identical:
differentiating both sides of that expression by the chain rule,

\dfrac{du}{dx} \times 2u = -1

, so

-\text{d}x = 2u \text{d}u

and the integral becomes

\displaystyle \int_{\sqrt{2}}^0 (2-u^2)u \times -2u \ \text{d}u

.

This is a polynomial expression of

u

so it's easy to evaluate.

Reply 8

studentFlipper

Original post by Smaug123

So the

u^2 = 2-x

substitution is essentially identical:
differentiating both sides of that expression by the chain rule,

\dfrac{du}{dx} \times 2u = -1

, so

-\text{d}x = 2u \text{d}u

and the integral becomes

\displaystyle \int_{\sqrt{2}}^0 (2-u^2)u \times -2u \ \text{d}u

.

This is a polynomial expression of

u

so it's easy to evaluate.

Aagghhh it's too complicated for me, I don't understand it. I know the essence of the chain rule,. I just don't see how it is applicable here, I can't get my head around it. I see that you differentiated both sides u^2 >> 2u and 2 - x >>> -1 but why you would multiply 2u by 2u/dx or why do we need a negative dx is what I don't get either. Usually when I am doing integrals like these, I am solving for dx or dt, and then plugging it into the integral. Even considering I do all that, I am still somewhat confused with how we got root 2 in there, this is all so complicated for me when it should be easy.

Original post by TenOfThem

As I said you have everything in root2s

Trying to grasp my head around it, too problematic to understand. I have everything in root of 2 as in my u's because they are rooted? You mean like turn 2u^3/2 into something like 2u times *root* u?

Do not change to decimal

Original post by davros

I haven't checked your working, but why have you converted everything to decimals?

You should know that

2^{3/2} = 2 \times 2^{1/2} = 2\sqrt{2}

and similarly

2^{5/2} = 2^2 \times 2^{1/2} = 4\sqrt{2}

so you can write everything as a multiple of

\sqrt{2}

Yes you are right! I should have used 1*(-du) instead. I'll try to remember that, I also know the powers multiplication conversion. The problem for is, that I think I wouldn't have thought of changing them to the roots in the first place, even when asked to give an exact value. I just wouldn't have thought of it. I don't know what I need, maybe mass practice... but as soon as I get to review exercises I forget half the stuff I tried to learn haha