Integral Problem...

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Overcore

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Oct 7, 2012
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This problem is giving me trouble. I'm just not sure where to go with it. I'd really appreciate if someone would walk me through the steps, since I'll have to know how to do this in the future.


Here's the integral I'm supposed to define:

Integral from 0 to a of (x^2)sqrt((x^3)+(a^3))dx



Having the two variables in the integral is throwing me off.
 
Overcore said:
This problem is giving me trouble. I'm just not sure where to go with it. I'd really appreciate if someone would walk me through the steps, since I'll have to know how to do this in the future.


Here's the integral I'm supposed to define:

Integral from 0 to a of (x^2)sqrt((x^3)+(a^3))dx



Having the two variables in the integral is throwing me off.
Click to expand...

Why do you think there are two variables? That 'a' is a constant!

If you cannot "see" the integral quickly - substitute

u = x3 + a3

du = 3*x2 dx

Now carry on ......
 
Overcore said:
(x^2)sqrt((x^3)+(a^3))
Click to expand...

Here are some technical pointers about notation.

The red grouping symbols are unnecessary because the Order of Operations already tells us to exponentiate before addition. In other words, those extra parentheses do not mathematically change anything.

The blue grouping symbols are okay, but you could leave them off also because a radical sign is itself a grouping symbol (i.e., we would evaluate the square root before multiplying by x^2).


Some people find the following expression easier to read. Cheers :cool:

x^2 sqrt(x^3 + a^3)
 
I've always erred on the side of more parentheses for some reason. I know they aren't necessary, but it helps me stay organized. But if everyone else prefers less parentheses, less it is!
 
As long as those are correctly placed, use as many parentheses as you feel comfortable. Absolute minimum is that clarifies order of operation.
 
So after I integrate I have

(2/9)u^(3/2)

which turns into

(2/9)(x^3+a^3)^(3/2)

Where do I go from here?
Do I just evaluate x from 0 to a and write out the following?


(2/9)(0^3+a^3)^(3/2)-(2/9)(a^3+a^3)^(3/2)

(2/9)a^(9/2)-(2/9)(2a^(9/2))

is that right?
 
Overcore said:
I know they aren't necessary, but it helps me stay organized.
Click to expand...

Ah -- I did not know that you knew. It's common for students who are beginning to learn software syntax to become confused over the use of parentheses, and thus go overboard.

But hey, if extra parentheses help you to stay organized, then please feel free to text things like sqrt(((((x^((((3))))))))+((((a^((((3)))))))))

heh, heh, heh -- just kidding, of course. :cool:

Seriously, you are free to use your own style -- regardless of other peoples' preference. My preference is to avoid extra symbols when texting (for readability).

Plus, I'm lazy.
 
Fair enough. I've had a few instances where I forgot them where it was important on tests, so now I put in extra just in case.
 
Overcore said:
So after I integrate I have

(2/9)u^(3/2)

which turns into

(2/9)(x^3+a^3)^(3/2)

Where do I go from here?
Do I just evaluate x from 0 to a and write out the following?


(2/9)(0^3+a^3)^(3/2)-(2/9)(a^3+a^3)^(3/2)

(2/9)a^(9/2)-(2/9)(2a^(9/2))

is that right?
Click to expand...

It looks good - but you can go one more step and simplify....
 
Is it


((2a^9/2)/9)(1-16sqrt2)?

And I get that a is just a constant, so that means I don't need a +C, right? Never want to forget about that +C...
 
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Overcore said:
2a^9/2
Click to expand...

In this case, we do need grouping symbols around the exponent. Agree?


I get that a is just a constant, so that means I don't need a +C, right?
Click to expand...

That's not correct.

(I do not think that you would ask this question, were you to understand why we add C.)

Integration results in a family of functions (a family with infinite members). These functions all behave exactly the same (thus they all have the same derivative), but their graphs are shifted vertically with respect to each other.

Adding constant C to a function results in a vertical shift. A known value of C determines the specific function in the infinite family.

The constant 'a' does not serve this purpose. To me, 'a' is a parameter.
 
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