Question

I am REALLY stuck on this one... This (I think) Definite Integral problem has me stuck at the first step, and I can usually at least get my "u" figured out... Would someone be willing to coach me through how to approach this problem, and solving it?

*I will post the question as the first comment below.

Any and all help is greatly appreciated!

Answer 1

Don't forget to show us YOUR efforts.

Answer 2

The problem:
\[\int\limits_{0}^{a}x \sqrt({x^2 + a^2})dx, a >0.\]

Answer 3

Strongly suggest that you consider a substitution. What would that subst. be?

Answer 4

I tried using (x^2 + a^2)^(1/2) as u, but got nowhere, and I can vaguely see some relationships in the book and WolframAlpha'a answers, but to be honest, the "a" is throwing me off, and I'm not sure where to start here... :/

Answer 5

...Is the best "u" going to be x^2?

Answer 6

Maybe just the stuff inside the radical? All of it?

Answer 7

No, but that's better than your previous venture.

Answer 8

Let u = x^2 + a^2.
Differentiate that, please. Find du/dx.

Answer 9

Notice that u=x^2+a^2 is all INSIDE the parentheses. Do not include the exponent in your substitution u (at least, not in this problem).

Answer 10

Hints: x usually represents a variable, whereas a usually represents a constant.

Answer 11

Gotcha! So...
\[u = x^2 + a^2\]
\[du = (2x + a^2)dx\]
?

Answer 12

Maybe \(u^{2} = x^{2} + a^{2}\)

Did you know you could do that?

Answer 13

Nooo! :D So... u = x+1?

Answer 14

Stop for a minute and think: why are we bothering with a substitution? What would be its purpose? Note that you have Radical (x^2+a^2), with x outside the radical.

Answer 15

Before we go any further, consider that a and a^2 are not variables, but constants, using the conventions that x usually represents a variable and a a constant.

Answer 16

Say it with me, now..."The derivative of a constant is ______."

Answer 17

To get a derivative that's in the integrand?

Answer 18

what is the derivative of a constant?

What is \[\frac{ d }{ dx }a^2?\]

Answer 19

Oh, duh Amon.. they're 0.

Answer 20

?

Answer 21

Correct. So, if your u is x^2+a^2, du/dx is what?

Answer 22

I thought you meant treat it like "e" or something.

Answer 23

Then...
\[du = 2x*dx\]

Answer 24

e without an exponent is another constant.
e with an exponent represents an exponential function.

Answer 25

Very good. du=2xdx. Note that you do have x outside the radical, but not the 2. Is there nevertheless a way to simplify the integral using substitutions?

Answer 26

Yes, I think? \[\frac{1}{2}du = x*dx\]

Answer 27

e - constant
e^2 - constant
e^x - exponential function

Answer 28

Then we can plug in the left for the right, and make what's in the radical "u".

Answer 29

Your original integral includes xdx, but not 2xdx. What are you gonna do about that? Good work ... you answered my qu as I was typing it.
So, replace your xdx with du/2. Under the integral, replace your x^2+a^2 with u.

Answer 30

What do you end up with?
Is it any easier to integrate than the previous form? Let's hope so.

Answer 31

Would appreciate your typing in the integral as it looks after 2 substitutions.

Answer 32

So given the algebra we did earlier working with "u"...
\[= \int\limits_{0}^{a}\frac{1}{2}\sqrt{u}*du\]
\[= \frac{1}{2}[\frac{2}{3}u^{3/2}] 0/a\]

Answer 33

\[= \frac{1}{3}(x^2 + a^2)^{3/2}, from 0 \to a\]

Answer 34

The limits of integration AFTER substitution are not necessarily the same as BEFORE substitution. Note that y ou let u=x^2+a^2.

Answer 35

But if I plug "u" back in, and am not plugging in the limit directly for the term "u" itself, then I keep the limits that were (are) in terms of "x" (or whatever the problem is initially in terms of), aye?

Answer 36

You can do that, but after going thru the trouble of substituting, why not keep u and forget about x^2+a^2? It's so much simpler.
If x goes from 0 to a, what happens to u? u is your new variable and you thus need to compute the appropriate limits on u.

Answer 37

There's more than one way to skin a cat. You can either STICK with u, or you could re-substitute x^2+a^2 for u. My preference is that you stick with u.

Answer 38

x varies from 0 to a. Thus, if u=x^2+a^2, u varies from (what?) to (what?)

Answer 39

The latter values are your new limits of integration./

Answer 40

Gotcha:) I will agree to respectfully disagree with you, as you are far more knowledgable than I, but I will plug x^2 + a^2 back in:)
Umm, u varies from... [2a^2 , a^2]
???

Answer 41

You can do that substitution; it's correct. However, by going back to x^2+a^2, you have abandoned u.

Answer 42

Do it both ways. You choose to plug x^2+a^2 back into the RESULT of your integration? Fine. Then let x vary from 0 to a.

You cvhoose to stick with u? Fine. Since u=x^2+a^2, and a varies from 0 to a, u varies from a^2 to 2a^2.

Answer 43

I admire hunny's suggestion, that you let u^2=x^2+a^2. Doing this will eliminate the square root operation, which is a good thing. Simplification!

Answer 44

Regardless of the method used, you MUST end up with the same result. Else something is wrong.

Answer 45

I'm with you there! I'm working it out now... :)

Answer 46

Okay, I got:
\[ = \frac{ 1 }{ 3 }(a^2)^{3/2}\] ??

Answer 47

Is that right?

Answer 48

Question: Can you simplify (a^2)^(3/2)?

making good progress.
I have to get off the 'Net right now, but will be back in 0.5-1.0 hour. Good luck!

Answer 49

Yes, you can, but is that the right answer in general, if I make (a^2)^(3/2) = a^(7/2)?

Answer 50

Let's re-write (a^2)^(3/2) as\[(a^2)^{\frac{ 3 }{ 2 }}\]

Answer 51

since one of the rules of exponentiaion is (x^a)^b=x^(ab), \[(a^2)^{\frac{ 3 }{ 2}}=\]

Answer 52

what?

Answer 53

Where is WolframAlpha getting the sqrt. of 2 here?

Answer 54

My bad on the exponents there...

Answer 55

If you use substitution, the integration becomes quite simple. I'd suggest you go thru with the integration yourself and then worry about that Sqrt(2) from Wolfram.

Answer 56

Here's what I got:
\[\frac{1}{3}[x^2 + a^2]^{3/2}, 0/a\]
\[= [\frac{1}{3}((a)^2+a^2)^{3/2}] - [\frac{1}{3}((0)^2+a^2)^{3/2}]\]
\[= [\frac{1}{3}(2a^2)^{3/2}] - [\frac{1}{3}(a^2)^{3/2}]\]
\[= [\frac{1}{3}(a^2)^{3/2}] = \frac{ 1 }{ 3 }a^{3}\]
How does that look? :/

Answer 57

Sorry: I was detracted by another posted problem.

Answer 58

Having taken a quick (not really careful) look at your work, my first impression is that you've done well.

Answer 59

Cool. Have to get off the 'Net pronto, but will be back later. Admire your persistence and hard work! See you later.

Answer 60

Do you guys know where the
\[2\sqrt{2} - 1 \]
comes from? :/

Answer 61

\[= [\frac{1}{3}(2a^2)^{3/2}] - [\frac{1}{3}(a^2)^{3/2}]\]
so you went from here to:
\[= [\frac{1}{3}(a^2)^{3/2}] = \frac{ 1 }{ 3 }a^{3}\]
how did you make the 2^(3/2) disappear?

Answer 62

\[=\frac{1}{3}2^\frac{3}{2}a^3-\frac{1}{3}a^3 \\ =\frac{1}{3}a^3(2^\frac{3}{2}-1)\]

Answer 63

I thought that the (2a^2)^(3/2) - (a^2)^(3/2) was just like subtracting 5x from 6x. So basically, the issue there was that I needed to distribute the power of 3/2 to the 2, and not treat it as such a coefficient?

Answer 64

you could have treated like if was outside the exponent if it was outside the exponent :p

Answer 65

\[2(a^2)^\frac{3}{2}-(a^2)^\frac{3}{2}=(2-1)(a^2)^\frac{3}{2}=1(a^2)^\frac{3}{2} \\ \text{ but } (2a^2)^\frac{3}{2}-(a^2)^\frac{3}{2}= 2^\frac{3}{2}(a^2)^\frac{3}{2}-(a^2)^\frac{3}{2}=(a^2)^\frac{3}{2}(2^\frac{3}{2}-1)\]

Answer 66

Ah.... I see.. :) Gotcha.

Answer 67

example
(2*5)^2 is not the same as 2*(5^2)

Answer 68

(2*5)^2=10^2=100
or you could say
(2*5)^2=2^2*5^2=4*25=100

but
2*(5^2)=2*25=50

Answer 69

2*(5^2) or 2*(5)^2 whatever

Answer 70

or you can even say 2*5^2 :p

Answer 71

2*(5^2)=2*(5)^2=2*5^2

but none of those are equal to (2*5)^2

Answer 72

So how would you treat or simplify this then:
\[[\frac{1}{3}(a^2 + a^2)^{3/2}] = [\frac{1}{3}(2a^2)^{3/2}] ????\]

Answer 73

That's where I pulled the two into play, but if it's not to be treated like a coefficient, I'm confused about what's going on. :?

Answer 74

the 3/2 is on the (2a^2)

Answer 75

\[\frac{1}{3}(2a^2)^\frac{3}{2} \\ \frac{1}{3}(2^1 a^2)^\frac{3}{2} \\ \frac{1}{3}(2^{1 \cdot \frac{3}{2}} a^{2 \cdot \frac{3}{2}})\]

Answer 76

right, so doesn't it distribute into the "^2", just as it would if it was for example (x^2)^(3/2)?

Answer 77

Okay, I see where you're going there, but that's just so not intuitive for me... My bad I guess!

Answer 78

\[a^{2 \cdot \frac{3}{2}}=a^3\]
since a>0

Answer 79

Right, I just didn't get the whole distributing the power across to the 2 in front of that a^2. :) I see what you mean completely now, albeit being still in a phase of working adjusting to thinking of everything this way. Thank you:)

Answer 80

oh ok
well you could just ignore 2 being raised to the 3/2 power

Answer 81

\[\text{ for example } 2^3=8 \text{ \not } 2 \]

Answer 82

I know that seems like a stupid example
\[(2a)^n=2^n a^n\]

Answer 83

Gotcha... gotcha indeed. I learn something new every night... :)

Answer 84

lol and you probably will for the rest of your life
we never stop learning well until we become a vegetable or dead

Answer 85

Good Lord, I better never stop learning!!!!

Answer 86

lol

Answer 87

good luck with living forever

Answer 88

;) I'll be here!