Understanding Integration: Unpacking the Formula and Solving Problems

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In summary, the conversation discusses the concept of integration and the confusion surrounding the formula \int u^n du = \frac{u^{n+1}}{n+1} + C. The speakers clarify that this formula is derived from the general rule that \frac{d}{dx}(u^n) = n u^{n-1} \frac{du}{dx}, and due to the constant of integration, the final answer may differ. They also discuss the importance of differentiating with respect to the correct variable. The conversation ends with a clarification on the use of abstract variables in solving integration problems.
  • #1
Ajoo
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Hi. I read a few things about integration on my maths book and googled it up. I think i got the idea of it, but there's something that's still confusing me:

I found this formula but i can't see how it can be true

[tex] \int u^n du = \frac{u^{n+1}}{n+1} + C [/tex]

however, as far as i remember:

(u^n)' = n*u'*u^(n-1)

How can this be right. where did that (u') go?

For example: please tell me how to solve this one:

[tex] \int (1-x^2)^\frac{1}{2} dx [/tex]

using the formula i'd say:

[tex] \frac{2(1-x^2)^\frac{3}{2}}{3} [/tex]

but this can't be right.:frown:
 
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  • #2
u' comes in when you differentiate with respect to a variable other than u. For example, if u=u(x), then:

[tex]\frac{d}{dx}(u^n) = n u^{n-1} \frac{du}{dx} [/tex]

If you differentiate with respect to u (in other words, take u=x in the above equation), then du/du=1, and so:

[tex]\frac{d}{du}(u^n) = n u^{n-1} [/tex]

Applying this to the indefinite integral above you get back the integrand as expected.

As for your second question, this is more complicated because u=1-x2, so du/dx is not 1 as it was before. The final answer will involve the arcsin function. If you're interested, this would go like:

[tex]\int \sqrt{1-x^2} dx = \int \frac{1-x^2}{\sqrt{1-x^2}} dx[/tex]
[tex]= \int \frac{dx}{\sqrt{1-x^2}} - \int \frac{x^2 dx}{\sqrt{1-x^2}} = \sin^{-1} x - \left( (x)(-\sqrt{1-x^2}) - \int (-\sqrt{1-x^2}) dx \right)[/tex]

[tex]\int \sqrt{1-x^2} dx = \frac{1}{2} (\sin^{-1} x + x\sqrt{1-x^2})[/tex]
 
  • #3
integrate

I think this is one nice example
 

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  • #4
Ajoo said:
(u^n)' = n*u'*u^(n-1)
is NOT correct. In fact, it really doesn't makes sense because it doesn't say with respect to what variable the derivative is done.

What is true is that
[tex]\frac{du^n}{dx}= n u^{n-1}\frac{du}{dx}[/tex]
In particular, if u= x,
[tex]\frac{du}{dx}= 1[/tex]
so
[tex]\frac{du^n}{du}= n u^{n-1}[/tex]

That also means that
[tex]\frac{1}{n+1}\frac{du^{n+1}}{du}= \frac{1}{n+1}(n+1)x^{n+1-1}= u^n[/tex]
so
[tex]\int u^n du= \frac{1}{n+1}u^{n+1}[/tex]
 
  • #5
It's simply...the abstractization of the antiderivative written in a very fancy way.

assume your problem asks for integration of 3x^2. you know answer is x^3.

So look. 3x^2 integrated => 3x ^ (2+1) divided by (2+1) => 3x^3 divided by 3 or...x^3.
 
  • #6
Robokapp said:
It's simply...the abstractization of the antiderivative written in a very fancy way.

assume your problem asks for integration of 3x^2. you know answer is x^3.

Hopefully, you know the answer is x^3+ C!:biggrin:
 
  • #7
not +C! but plain +C w00t!
But yeah, good point. Just for the record, in first few months of calculus I forgot every dx and every +C I encountered and should have written...all until differentials. Then...God, they mattered :D
 
  • #8
Robokapp said:
It's simply...the abstractization of the antiderivative written in a very fancy way.

assume your problem asks for integration of 3x^2. you know answer is x^3.

So look. 3x^2 integrated => 3x ^ (2+1) divided by (2+1) => 3x^3 divided by 3 or...x^3.
"abstractization"?
 
  • #9
Maybe i used the wrong word, I'm not a native english speaker...The thing where universal variables replace numbers...basically a model that you can use for your numbers if you can figure out what goes where. At times I found it to be more difficult than just...say it in baby terms.

Oh you know what I mean! I was trying to sound smart :(
 
  • #10
Robokapp: "Abstraction" is a perfectly good word and works in place of "Abstractization". Unfortunately, in English, it is easy to create new, overly long, words by adding "endings". Don't use long words where shorter ones will do!

By the way, your English is far better than my (put pretty much any language you wish here).
 
  • #11
Abstraction...

Abstract is the root word...Abstraction is the noun formed by derivation...Abstractization is the noun reflecting "Abstraction done by someone on something"...a noun coming from the verb...which I think goes "to abstractize". Or is it "to abstract"?

Well that's how i thought it out I think. I don't start mubling these things before i write a word...it just comes out.

I'll remember it for the future hehe. Thx!
 

FAQ: Understanding Integration: Unpacking the Formula and Solving Problems

What is integration?

Integration is a mathematical process in which we find the area under a curve or the accumulation of a quantity over a certain interval.

How is integration different from differentiation?

Integration and differentiation are two fundamental operations in calculus. While integration finds the area under a curve, differentiation finds the slope of a curve at a given point.

What are the different types of integration?

The two main types of integration are indefinite integration, which gives the general solution to an integral, and definite integration, which gives a specific numerical value for an integral over a given interval.

What are some common integration techniques?

Some common integration techniques include substitution, integration by parts, and trigonometric substitution. These techniques are used to simplify integrals and make them easier to solve.

What are some real-world applications of integration?

Integration has many real-world applications, such as calculating the area under a velocity-time graph to find the distance traveled, finding the volume of irregularly shaped objects, and determining the amount of work done by a changing force. It is also extensively used in fields such as physics, engineering, and economics.

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