Integrals are harder than derivatives, why?

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I understand the concept of derivatives but when it comes to integrals and their uses I do not understand what they do and where you use them.In derivatives you can understand how a function changes but in integration everything is so illogical.Can someone explain me the use of integrals in calculus?I mean all I could understand is that there is some +C which is a constant but I have no idea where that come from.What does this +C even mean?When I look at derivatives I can see that the function changes but when I look at a integral I have no idea what a function does in that specific function.All I know is that I learned(more memorized) and I couldn't understand the complexity of them.
I have a math book full of exercises and it doesn't explain at all how a integral works.It just shows me some integrals that I learned in highschool and most of them don't even show the proof behind them.
 

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  • #2
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It is just the opposite direction.
If f'(x) is the derivative of f(x), then f(x) is an integral function of f'(x).

You already encountered cases of this much earlier: If 5+6=11, then 11-6=5. Subtracting 6 is the inverse action of adding 6. If 3*4=12, then 12/4=3. Dividing by 4 is the inverse of multiplying by 4.

If you add a constant to a function, its derivative does not change: f(x) and f(x)+3 have the same derivative. If you are only given the derivative, you don't know if it is the derivative of f(x), of f(x)+6423, or something else. Therefore the +c is added to keep all options. Every value of c leads to a possible integral function.
 
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  • #3
Stephen Tashi
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when it comes to integrals and their uses I do not understand what they do and where you use them.
.
Begin by making a distinction between "antiderivative" and "definite integral". Some people use the term "integral" to refer to both those concepts, but the two concepts have different definitions. Informally, a "definite integral" of a function refers to an area "under" its graph. The antiderivative of a function f(x) is another function F(x) such that F'(x) = f(x). For example, if f(x) = 2x, then both F_a(x) = x^2 and F_b(x) = x^2 + 6 are antiderivatives of f(x).

The connection between definite integrals and antiderivatives is given by The Fundmental Theorem of Calculus, which you should know if you have taken a calculus course. It's normal for a student not to have an intuitive understanding of why The Fundamental Theorem of Calculus should be true, but there's no excuse for not knowing what that theorem says and how it connects the idea of definite integral with the idea of antiderivative. Do you understand what the Fundamental Theorem of Calculus says?
 
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  • #4
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That is not what I asked.I asked why is integration harder than differentiation.I said that in derivatives the function of f(x) changes over time df(x)/dx.That is easy to understand but understanding why the integral of some functions are the derivatives of that function is not.
 
  • #5
Stephen Tashi
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the integral of some functions are the derivatives of that function is not.
You probably mean "...are the antiderivatives...".

You didn't say whether you know what the Fundamental Theorem Of Calculus says.
 
  • #6
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The truth is that you can derivate functions but not all functions are the derivative of a function.How does that make sense?For ex the integral of e^x/x has no derivative.Now I'm really confused.I just don't see the logic of integrals in derivatives.Can you explain me that?
 
  • #7
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That is not what I asked.I asked why is integration harder than differentiation.I said that in derivatives the function of f(x) changes over time df(x)/dx.That is easy to understand but understanding why the integral of some functions are the derivatives of that function is not.
If you dig to the bottom of this question, the answer will be: because multiplication is harder than addition.
To build a derivative, we attach a ruler at some point of the object, which is a linear, i.e.an additive thing. To integrate, we have to calculate a volume, which is a multiplication, or many small ones in this case. The formula ##e^x \cdot e^y = e^{x+y}## illustrates the difficulty: the derivative corresponds to the addition in the exponent, while the integration is the multiplication on the base line. That's why most differential equation are solved by searching for something like ##c(x)e^{a(x)}##.
 
  • #8
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I do understand calculus but when it comes to integrals the theory doesn't make any sense.In differentiation you use product rules as an exemple but in integration by parts you don't.How does that even work?
 
  • #9
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I do understand calculus but when it comes to integrals the theory doesn't make any sense.In differentiation you use product rules as an exemple but in integration by parts you don't.How does that even work?
Differentiation for many functions is very straightforward -- you use the product rule, chain rule, etc., and you might have to use several rules in succession in a certain order. Going backward (antidifferentiating) is much less straightforward, since the procedure is much less "cookbook" than differentiation.

Whether you know it or not, integration by parts is the reverse operation of the product rule. If h(x) = f(x) * g(x), then h'(x) = f(x) * g'(x) + f'(x) g(x). This equation can be transformed to f(x) * g'(x) = h'(x) - f'(x) * g(x). If you multiply both sides by dx and antidifferentiate both sides you get ##\int f(x) g'(x) dx = h(x) - \int f'(x) g(x) dx##, which is integration by parts.
 
  • #10
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The truth is that you can derivate functions
In English we say you can differentiate functions.
Derivate is a word, but not one that's used in discussions of calculus.
 
  • #11
Stephen Tashi
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The truth is that you can derivate functions but not all functions are the derivative of a function.
Have you taken a calculus course? You seem unfamiliar with standard terminology. We "differentiate" functions , not "derivate" them.

For ex the integral of e^x/x has no derivative.
If you understood the concepts of integral and derivative, you'd realize you haven't given a specific example.

Now I'm really confused.I just don't see the logic of integrals in derivatives.Can you explain me that?
I have to guess at your level of understanding. The fact that you are carless with terminology - such as writing "derivative" when you mean "antiderivative" suggests that your are approaching calculus as set of procedures for manipulating symbols - i.e. as a more sophisticated version of manipulations that are done in elementary algebra. You appear frustrated because the manipulations permitted for antidifferentiation are more complicated that those permitted for differentiation. You won't fully understand why certain procedures of manipulating symbols are pemitted unless you understand the verbal concepts that underlie the manipulations. If you are using a standard calculus text, there are verbal explanations for the concepts of definite integral and antiderivative. So my guess is that you are ignoring the explanations in your text materials. People on this forum can present their own versions of what's aleady in your textbooks and perhaps the personal attention will motivate you to pay closer attention to these ideas.
 
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  • #12
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In English we say you can differentiate functions.
Derivate is a word, but not one that's used in discussions of calculus.
I struggle between derivative and derivation. It's the same at its core, but then again different in context.
 
  • #13
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Sorry I meant differentiate.
 
  • #14
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I struggle between derivative and derivation. It's the same at its core, but then again different in context.
I'm sure it's confusing, but then English is not known for being logical.

As far as derivative and derivation, it's just something you have to remember, since the usage isn't logical at all.

Differentiation is the operation used to calculate a derivative. Or, you can differentiate a function to get its derivative, but you don't derive a function to get its derivative.

The word derivate is a noun, so it's not an action word (you can't derivate something). From Oxford Dictionaries, "something derived, especially a product obtained chemically from a raw material." I've never seen "derivate" used in any calculus textbook, at least any written by English speakers.

You can start with a quadratic equation, and use completing the square to derive the Quadratic Formula. Or, the English word "cotton" is derived from the Spanish word "algodon," which is almost identical to the Arabic word "al godon."
 
  • #15
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While it may be true what you say but when it comes to integration from -inf to +inf then integrals can be hard.I mean even in limits you can solve that but can you solve integrals that tend from -inf to 0 and from 0 to +inf?Is that even possible?
 
  • #16
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Sorry I meant differentiate.
No problem and no need for an apology. I wasn't trying to be critical. I know you're not a native speaker of English. I was just trying to help you out.
 
  • #17
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What I meant in the thread is that I want to find the complexity of integrals over derivatives.If you can look at integrals that tend to inf just like limits you can see that integrals can't be solved like limits.Another fact that I do not understand is why a limit is placed in a solution of a integral.At least in limits I remember that you couldn't use such a rule like this one.
 
  • #18
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Uh no one knows?
 
  • #19
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I'm sure it's confusing, but then English is not known for being logical.
No that isn't it. I've studied a lot about Lie algebras, and they have derivations, which are linear mappings defined by ##d([a,b])=[d(a),b]+[a,d(b)]##. This is just a version of the Leibniz rule, as the Jacobi identity is, because it is deduced - or should I say derived - from derivatives. It's basically the same thing, just not calculus. And as I'm more used to write derivation, it occasionally slips into derivatives. It would be equally difficult in German, but the German word for derivative is "Ableitung". It's the literal translation of the Latin one, so it's a bit easier because of that.
 
  • #21
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While it may be true what you say but when it comes to integration from -inf to +inf then integrals can be hard.I mean even in limits you can solve that but can you solve integrals that tend from -inf to 0 and from 0 to +inf?Is that even possible?
Sure, these kinds of definite integrals come up fairly often. For example, ##\int_0^\infty e^{-x}dx## is a fairly simple example of a convergent improper integral whose value is 1.
In contrast, the improper integral ##\int_{-\infty}^0 e^{-x}dx## diverges.
 
  • #22
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But how can someone use limits in integrals(antidifferentiation functions)?I can understand that in limits but not in integrals.I thought that integrals can't have limits because that would make no sense and break the rules of calculus.
 
  • #23
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But how can someone use limits in integrals(antidifferentiation functions)?I can understand that in limits but not in integrals.I thought that integrals can't have limits because that would make no sense and break the rules of calculus.
I don't quite understand. Integrals are limits. The limit of the sums of tiny rectangular volume elements needed to fill a curved area. The tinier the rectangles, the better the approximation. That's where integrals came from.
 
  • #24
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There are many math problems that become a lot more difficult when turned into inverse problems. As an example not related to calculus, think about finding the prime factors of a large integer, compared to multiplying known prime factors to get a large number.
 
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  • #25
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Regarding the use of integrals:
Some real-world examples may be helpful. The integral of a function is the accumulation of the function values over a range of its inputs. If you know the velocity of a car as a function of time, v(t), then you can integrate it to keep track of the position of a car whose initial position at time 0 is ##p_0##, ##p(t) = p_0 + \int_0^t v(s)ds##. (You would need to do that for every dimension that the car can travel in.) One step further, you can accumulate the accelerations of a car to keep track of its velocity. This is a very common application because accelerometers are very easy to include in a vehicle.

Regarding the relative difficulty of determining a closed-form equation for an integral:
A function with a derivative at a point is very well behaved at that point and around it. The derivative is only a local property and the behavior of the function some distance from that point are irrelevant. The function will be continuous at the point where it has a derivative.
Very bazarre functions have integrals. The functions may be extremely discontinuous. Furthermore, since the integral is an accumulation of the function values over a range of input values, it depends on the behavior of the function over the entire range. That is a much more difficult thing to handle than the local property of a derivative.
 
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