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How can we prove the Integral that inverse of the derivative I am trying to say this one, for $f\left( x\right) =x$ and,

$\int \left( xdx\right) =\dfrac {x^{2}} {2}$ and in derivative,

$f^{'}\left( x\right) =1 $ but why? someone can prove? that are pass the pages that has prove I'm bloody curious on it

beingmathematician
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    See this question and its top answer. – Bobson Dugnutt May 03 '17 at 15:38
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    Could you please reformulate clearly your question? I'm afraid it is just about the basics of derivatives and integrals, in this case you should first search by yourself a bit more... – MysteryGuy May 03 '17 at 15:40
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    Newton's proof that the derivative of the integral is the original function is pretty easy to follow: http://www.maa.org/press/periodicals/convergence/teaching-the-fundamental-theorem-of-calculus-a-historical-reflection-newtons-proof-of-the-ftc – Chappers May 03 '17 at 15:43
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    You just need the first paragraph of that top answer. Also, note that $\int^x \left(\frac{d}{dt} t\right) dt = x + C$ but $\frac{d}{dx} \int^x tdt = x$. – Paichu May 03 '17 at 15:44

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The second question involves less; so I begin with it. From the definition of differentiation, for every $x$ we have $$ \frac{f(x+h) - f(x)}{h} = \frac{x+h-x}{h} = 1 \to 1 $$ as $h \to 0$. So $f'(x) = 1$ for all $x$.

There are three concepts of integrals in calculus of functions $\mathbb{R} \to \mathbb{R}$: definite integrals, indefinite integrals, and primitives. (For the sake of illustration my use of terms here may not be that proper regarding the modern standard; however, one will be not be confused if he looks at the according definitions.)

The partition-limiting stuff defines the concept of the definite integral $\int_{a}^{b}f$ of a (suitable) function $f$ over $[a,b]$. If $c \in [a,b]$ and if $\int_{c}^{x}f$ exists for every $x \in [a,b]$, then the function $x \mapsto \int_{c}^{x}f$ is then called the indefinite integral of $f$. Now we have been still in the realm of partition-limiting construction. A primitive of $f$ is defined in terms of differentiation; i.e. a function $F$ is called a primitive of $f$ on $[a,b]$ if $F' = f$ on $[a,b]$. Saying "a primitive" instead of "the primitive" is due to that ($F +$ constant)$' = f$. A usual mnemonic notation of a primitive of $f$ is simply $\int f$. (Don't worry about the $dx$ thing; it makes no difference to let it appear in the present case.) By the way, the fundamental theorem of calculus is a result asserting that if $f$ is continuous on $[a,b]$ then the indefinite integral of $f$ is a primitive of $f$ on $[a,b]$! Formally, if $f$ is continuous on $[a,b]$ then $(\int_{a}^{x}f)' = f$ on $[a,b]$. It follows from that $\int_{a}^{b}f = \int_{a}^{c}f + \int_{c}^{b}f = -\int_{c}^{a}f + \int_{c}^{b}f = (\int f)(b) - (\int f)(a)$ the so-called second fundamental theorem of calculus. (see the definition of primitive.) Regarding your first question, the equality says that $x \mapsto x^{2}/2$ is a primitive of $x \mapsto x$, which can be seen by $(x^{2}/2)' = x$ by chain rule. Note again that the primitives themselves involve only differentiation; it is the fundamental theorem of calculus that turns many of the problem of finding definite integrals into the problem of finding primitives.

Yes
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