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In the wikipedia article shown below, it says $A(x)$ represents the area beneath the curve $y=f(x)$ between $0$ and $x$. enter image description here

Then it says $f(x)=A'(x)$, i.e. $A(x)=\int f(x)dx$

which means $\int f(x)dx$ represents the area beneath the curve $y=f(x)$ between $0$ and $x$.

I think there is nothing wrong up to here.


Now if we put $f(x)=e^{x}$ then $A(x)=e^{x}$ and at $x=0$, $A(x)=e^{0}=1$

That is, the area $A(x)$ beneath the curve $y=f(x)$ between ($0$ and $x=0$) is $1$.

Now how can we get non-zero area when we find the area between same points (i.e. $0$ and $0$)

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    $\begingroup$ For the area on $[a,b]$, you need $A(b)-A(a)$ (assuming $f$ is positive, as $e^x$ is) so for "$[0,0]$" you don't want $A(0)$, but $A(0)-A(0)$, which is of course $0$. $\endgroup$ – StackTD May 2 '17 at 8:24
  • $\begingroup$ But as said in wikipedia, doesn't $A(x)$ in general means area from 0 to x $\endgroup$ – Joe May 2 '17 at 8:25
  • $\begingroup$ What do you mean "from $A$"? $A$ is not a (starting) point. $\endgroup$ – StackTD May 2 '17 at 8:27
  • $\begingroup$ I have edited the comment $\endgroup$ – Joe May 2 '17 at 8:28
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    $\begingroup$ Yet another situation where you get in trouble if you forget the "$+ c$" part of an antiderivative. $\endgroup$ – MartianInvader May 2 '17 at 21:09
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This is because your definition of $A$ is not quite right. For $f(x) = e^x$, $A(x) = e^x - 1$. We can see this matches with the properties you've specified, because $A'(x) = e^x = f(x)$. Moreover, $A(0) = e^0-1 = 0$ as required. Now you're probably thinking "where did the $-1$ come from?"

Consider an area from $x=a$ to $x=b$. Then it's equal to the area from $0$ to $b$ minus the area from $0$ to $a$, right? So the area from $a$ to $b$ is given by $A(b)-A(a)$. For our function, this is $(e^b-1)-(e^a-1) = e^b-e^a$. The $1$s have disappeared! This shows that the $1$ only matters for the absolute case, but is irrelevant in relative cases. Why does it matter in the absolute case? This is because your "$0$" could have been defined anywhere. No matter where you move $0$, the value of $A(b)-A(a)$ won't change, but clearly $A(x)$ will. Now you can see that the $-1$ is actually just a translation factor, determining where "$0$" is. In fact, what you did was a calculation to find that factor.

"I know the area $A(0)$ must be zero, but I also know that $A'(x) = e^x$. Therefore I can conclude that $A(x) = e^x-1$."

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  • $\begingroup$ English is not my native language, hence the question: is $-1$ a (scaling) factor in this case? Wouldn't "addend" be a better word? $\endgroup$ – Kamil Maciorowski May 2 '17 at 19:36
  • $\begingroup$ Yes you're right, scaling factor doesn't sound quite right because I couldn't think of anything better at the time. I'll see if I can think of something else now. $\endgroup$ – Harambe May 2 '17 at 22:41
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    $\begingroup$ It looks like the word you are looking for is "offset" $\endgroup$ – Rad80 May 3 '17 at 11:56
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Now if we put $f(x)=e^{x}$ then $A(x)=e^{x}$ and at $x=0$, $A(x)=e^{0}=1$

That is, the area $A(x)$ beneath the curve $y=f(x)$ between ($0$ and $x=0$) is $1$.

Now how can we get non-zero area when we find the area between same points (i.e. $0$ and $0$)

Why do you assume $A(x)=e^x$ if $f(x)=e^x$? Probably because $(e^x)'=e^x$ but the same is true for $(e^x+c)'=e^x$ , for any $c \in \mathbb{R}$.

If $A(x)$ is introduced as the area under $f$ (assuming $f$ is positive) on the interval $[0,x]$, then for $f(x)=e^x$ you are right that $A(x) \ne e^x$; rather: $$A(x) = \int_0^x e^t \, \mbox{d}t = e^x-1$$ This agrees with what you expect for $A(0)$ since now $A(0) = e^0-1 = 0$.

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The fundamental theorem of calculus says that, if $f$ is a continuous function on some interval $I$ and $a\in I$, then $$ F(x)=\int_{a}^x f(t)\,dt $$ (for $x\in I$) is an antiderivative of $f$.

The function $A(x)=e^x$ is an antiderivative of $f(x)=e^x$ as well.

Note the emphasized “an”: there is no unique determination of the antiderivative.

However, by a well known consequence of the mean value theorem, the functions $A(x)$ and $F(x)$ differ by a constant. What is this constant? Easy: since $A(x)=F(x)+c$, for some constant $c$, $$ A(a)=F(a)+c $$ and by definition $F(a)=0$, so $c=A(a)$. Therefore $$ F(x)=A(x)-A(a) $$ and so $$ F(x)=e^x-e^a $$ In the particular case of $a=0$, you get $$ F(x)=\int_0^x e^t\,dt = e^x-e^0=e^x-1 $$ and, for $x=0$, we have $F(0)=0$, as expected.

More generally, if $A(x)$ is an antiderivative of $f(x)$, then $$ \int_a^b f(t)\,dt = A(b)-A(a) $$

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Note that $f(x)=A'(x)$ does not follow $A(x)= \int f(x) d x$ since in here $A(x)$ is a specific function but $\int f(x) dx$ gives all the possible antiderivatives of $f$. For example, if $f(x)=2x$ then $\int f(x) dx=x^2+c$ for any $c$, where as $A(x)$ could be $x^2+1$ but cannot be $x^2+c$ for any $c \ne 1$.

Hence, from $f(x)=A'(x)$, you can only follow that if $F(x)$ is one of the antiderivative of $f$ then $$A(x)=F(x)+c=e^x+c. \tag{1}$$

Thus, you can't imply to $A(x)=e^x$ from $f(x)=A'(x)$. Instead, it is automatically understood that area between $0$ and $0$ is $0$, so that means $A(0)=0$. Hence, from (1) we follow $c=1$. Thus, $A(x)=e^x-1$ is the area between $0$ and $x$.

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    $\begingroup$ +1 I think you have identified the real cause of the OP's problem: thinking of the indefinite integral as a function rather than a set of functions. Others have said this implicitly, but you're the one who tracked the problem to the notation. I wish he'd accepted this one instead. $\endgroup$ – Ethan Bolker May 2 '17 at 22:56
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The problem you are having might be the following: first, let us denote the antiderivative of $f$ by $A$, and the area of $f$ from $0$ to $x$ by $S$. Then one should have $S=\int_0^x fdx$, but not $A(x)=\int_0^xfdx$. In fact, one can only obtain that $A(x)=\int_0^xfdx+A(0)=S+A(0)$.

Now consider your example, you have your $A(x)=f(x)=e^x$, then your area should be $A(x)-A(0)=e^x-1$. Inserting $x=0$ you find the area is $0$.

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  • $\begingroup$ This might be confusing because you've defined $A(x)$ to be the area under the function for $(-\infty, x]$ rather than $[0,x]$ $\endgroup$ – Harambe May 2 '17 at 8:42
  • $\begingroup$ An antiderivative needs not to have an integral form due to the definition. $\endgroup$ – ehochix May 2 '17 at 8:46
  • $\begingroup$ The antiderivative? $\endgroup$ – Carsten S May 2 '17 at 15:37

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