Definite integral

Evaluate

\int^5_3 \sqrt {x + 2\sqrt {2x - 4}} + \sqrt {x - 2\sqrt {2x - 4}}\ dx

The options given are:
A)

2(\sqrt 5 + 1/5)

B)

4 (\sqrt3-1/3)

C)

3(\sqrt2-1/2)

D)

5(\sqrt 6+1/6)

E) None of these

Using the substitution

u=2x-4

I get the answer as B). However 2 other people get

4 \sqrt {3} - \frac {2}{3}\sqrt {2}

ie E) None of these.

Which is correct?

Scroll to see replies

The other two are right I think :smile:

:smile:

OP

rnd

The other two are right I think :smile:

:smile:

Your reason being?

Sub in u^2 = 2x - 4 - the square roots disappear easily that way.

Study Forum Helper

Thomas,

The two other people are right because I just used a computer to calculate the area. I will work through it myself shortly but am just preparing some work for tomorrow first.

OP

Swayum

Sub in u^2 = 2x - 4 - the square roots disappear easily that way.

Actually I have a solution. I'm more interested in knowing the correct answer.

alexmahone

Actually I have a solution. I'm more interested in knowing the correct answer.

You don't have a solution if it doesn't give you the right answer

3 of us on TSR agree with those other people

OP

Mr M

Thomas,

The two other people are right because I just used a computer to calculate the area. I will work through it myself shortly but am just preparing some work for tomorrow first.

Thank you! :smile:

:smile:

But it is usually unlikely that the answer to an olympiad problem is "None of these".

OP

Could any of you please post the full solution leading to answer E) ?

I'd really appreciate it.

Swayum's hint is all you need.

He's too kind. :smile:

:smile:

Mr M, are you a university profesoor, or an undergraduate student? Just wandering.

OP

rnd

Swayum's hint is all you need.

He's too kind. :smile:

:smile:

But my substitution of

u=2x-4

is equally good!

alexmahone

But my substitution of

u=2x-4

is equally good!

You got the wrong answer. Go with the one suggested

OP

SimonM

You got the wrong answer. Go with the one suggested

Why don't you try the substitution

u=2x-4

? Then the integrand reduces to

\sqrt{2u}

.

OP

Simon, in case it interests you, there's a heated debate going on about this problem on MathLinks; you might want to take a look.

Study Forum Helper

DeanK2

Mr M, are you a university profesoor, or an undergraduate student? Just wandering.

Alex's substitution works by the way. Just doing it.

I'm a maths teacher.

Sketching it, I still get the mathematica solution

alexmahone

Why don't you try the substitution

u=2x-4

? Then the integrand reduces to

\sqrt{2u}

.

Should that be

\frac{\sqrt{u}}{\sqrt{2}}

?

OP

rnd

Should that be

\frac{\sqrt{u}}{\sqrt{2}}

?

The integrand reduces to

(\frac{\sqrt{u}}{\sqrt{2}}+\sqrt2)+(\frac{\sqrt{u}}{\sqrt2}-\sqrt2)=\sqrt{2u}=2\sqrt {x-2}

And what about dx/du? (it's 2/root2 - 2/root2 by the way).

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