## Derivatives and Integrals

What I have learnt in school is that differentiation and integration are opposites.

By integrating a function we find the area under the graph. So, integration gives us the area. Differntiation gives slope of the function.

If I am right by saying differentiation and integration are opposites, then is area under the graph the opposite of the slope of the graph?
 Recognitions: Gold Member Homework Help Science Advisor What dioes "opposite" mean?? Differentiation UNDO what indefinite integration does to a function f(x), that is: Diff(Int(f(x))=f(x), whereas indefinite integration UNDO, up to an arbitrary error constant, what differentiation did to the function f(x). that is: Int(Diff(f(x))=f(x)+some constant.

 Quote by arildno Differentiation UNDO what indefinite integration does to a function f(x), that is: Diff(Int(f(x))=f(x), whereas indefinite integration UNDO, up to an arbitrary error constant, what differentiation did to the function f(x). that is: Int(Diff(f(x))=f(x)+some constant.
Pardon me if I sound stupid.
If we undo area under the graph, we get slope of that function then? Now I don't know how to undo the area or if thats even possible. Enlighten me.

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## Derivatives and Integrals

 Quote by Swetasuria Pardon me if I sound stupid. If we undo area under the graph, we get slope of that function then?
no.
read what I wrote,
 Mentor Swetsuria, Notice that arildno said indefinite integration (no limits of integration). To get the area under the graph of a function, you use a definite integral. The Fundamental Theorem of Calculus state in one of its two parts that differentiation and integration are essentially inverse operations. $$\frac{d}{dx}\int_a^x f(t) dt = f(x)$$ Although the integral in this formula is a definite integral, due to the fact that one of the limits of integration is a variable (x), the integral represents a function of x.
 Recognitions: Gold Member Homework Help Science Advisor What I did write, however, is closest precise interpretation of the vague and useless "opposite"-concept OP started out with.. That's why I wrote it like that. OP is far too confused at present to tackle ideas of "area under of a graph" and "slope of the function". The precisest meaning of "oppositeness" needs to be understood by him first.
 When you subtract and add the same number, the two operations undo each other. When you multiply and divide by the same number, the two operations undo each other. So would you consider addition and subtraction to be opposites? Would you consider multiplication and division to be opposites? Well, how come? The only simple answer is that they undo each other. Well, indefinite integration and differentiation undo each other, so if that is your definition of "opposites" then yes, by all means consider them opposites. However, the word "opposites" is variable in meaning between different people, so maybe you should rephrase your question as: "What is the intuitive reasoning behind differentiation and indefinite integration undoing each other?" Well, by definition, indefinite integration is just the reverse of differentiation, so that would be your answer. A better question would be: "Why does indefinite integration represent the rate of change of the area underneath a function?" Well, this question is one that I've wondered many times myself. I'll let you know if I find out

 Quote by Mark44 Swetsuria, Notice that arildno said indefinite integration (no limits of integration). To get the area under the graph of a function, you use a definite integral. The Fundamental Theorem of Calculus state in one of its two parts that differentiation and integration are essentially inverse operations. $$\frac{d}{dx}\int_a^x f(t) dt = f(x)$$ Although the integral in this formula is a definite integral, due to the fact that one of the limits of integration is a variable (x), the integral represents a function of x.
Okay, so is there a relation betwwen the slope of curve and area under curve (if there is one, I mean)? I'm pretty confused here.
 This PDF has a great mathematical proof of the fundamental theorem of calculus on pg.61 http://www.matematica.net/portal/e-b...ngineering.pdf
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