Solving Basic Integrals: A Tutorial

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Homework Help Overview

The discussion revolves around understanding the process of integration, particularly focusing on definite integrals and the relationship between differentiation and integration. The original poster expresses confusion about how to approach an integral problem involving a function F(x).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the inverse relationship between differentiation and integration, with some attempting to clarify the process of finding antiderivatives. Questions arise about the correct application of integration techniques and the proper setup for definite integrals.

Discussion Status

The conversation includes attempts to clarify misunderstandings about integration and differentiation. Some participants provide guidance on finding antiderivatives, while others question the original poster's approach and assumptions. There appears to be an ongoing exploration of the correct method without a clear consensus.

Contextual Notes

Participants note specific errors in the original poster's understanding of derivatives and integrals, highlighting the need for careful attention to the relationships between these operations. The original poster is preparing for a quiz, which adds urgency to the discussion.

lemonlee
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Homework Statement


I've been trying to find a website to explain this to me, but I'm still confused as to how to do an integral. The one that I have been trying to figure out is this:

Integral from 0 to d of F(x)dx

i can't figure out where the integral sign is either...sorry! but 0 is on the bottom and d on the top


Homework Equations



I know that it's almost the opposite of deriving an equation, d/dx

The Attempt at a Solution


no idea! I need to figure this out for a quiz on thursday and my professor just confuses me more when i try to get help.
 
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The operations of differentiation and integration (or antidifferentiation) are pretty much inverse operations.

If you know, for example that
[tex]\frac{d}{dx} x^3 = 3x^2[/tex] then you know an antidifferentiation formula,
[tex]\int{ 3x^2 dx} = x^3 + C[/tex]

or equivalently,
[tex]\int{ x^2 dx} = \frac{1}{3}x^3 + C[/tex]

If you're working with a definite integral (with limits of integration), you still need to find an antiderivative, as you do with an indefinite integral like the ones above in my post. The only difference is that you evaluate the antiderivative at upper limit and lower limit and then subtract the latter from the former.


One thing you said, "opposite of deriving an equation, d/dx" is incorrect. You don't apply the d/dx operator to an equation; you apply it to a function.
 
so if the equation that I'm trying to intergrate is F=-kx^4
i first have to find the derivative of -kx^4=4x^3 (right?) and then

4x^3dx=1/4x4+C

and from there i put in the two values (d and 0) and subtract them to find the definite integral?
 
NO, because the derivative of (1/4)x4+ C is (1/4)(4 x3)= x3, not 4x3. You need to multiply your result by -4: [itex]\int -k x^3 dx= -k \int x^3dx= -4k x^4+ C[/itex].
 
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lemonlee said:
so if the equation that I'm trying to intergrate is F=-kx^4
i first have to find the derivative of -kx^4=4x^3 (right?) and then

4x^3dx=1/4x4+C

and from there i put in the two values (d and 0) and subtract them to find the definite integral?

No, what you have to do is find a function whose derivative is -kx^4. It will be some multiple of x^5.
 
HallsofIvy said:
NO, because the derivative of (1/4)x4+ C is (1/4)(4 x3= x3, not 4x3. You need to multiply your result by -4: [itex]\int -k x^4 dx= -k \int x^4dx= -4k x^4+ C[/itex].
Halls, the last expression on the right is incorrect. It should be [tex]-k \frac {x^5}{5} + C[/tex]


lemonlee,
The check for the above is to differentiate the result gotten by antidifferentiation.

[tex]\frac{d}{dx}(-k \frac {x^5}{5} + C )= -k \frac{5x^4}{5} + 0 = -kx^4[/tex]
 
Right- his answer had me integrating the wrong thing!
 

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