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I have recently started studying calculus and I constantly find my self baffled by its notations. Specifically can somebody explain the difference between:

$$\int_{a}^{b} x^2 dx$$ $$\int_{}x^2 $$ $$\int_{a}^{b} x^2$$

and $$\frac{d}{dt}x^2$$ $$\frac{dx}{dt}x^2$$

3 Answers 3

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Disclaimer: These are my interpretations of the notation, but other people may have other ways of writing things.

The first step is noting that

$$\int_a^bf(x)\ dx=\text{Area}$$

while

$$\int f(x)\ dx=\text{anti-derivative}$$

These are two separate concepts, bound together by the fundamental theorem of calculus:

$$\int_a^bf(x)\ dx=F(b)-F(a),\text{ where }F(x)=\int f(x)\ dx$$

Anti-derivatives are defined as follows:

$$\frac d{dx}F(x)=f(x)\implies F(x)=\int f(x)\ dx$$

It is, lastly, in my greatest opinion that $\int f(x)$ is unclear notation. A lack of that final $dx$ will cause you great misfortune, for example, what does the following mean?

$$\iint\frac xy\sin(z)=?$$

Does it mean to integrate with respect to $x,y$ or $z$? And in which order? Clearly, this expression makes no sense, but,

$$\iint\frac xy\sin(z)\ dx\ dz$$

Makes perfect sense.


Derivatives, on the other hand, have their own tricky notations:

The following means to take the derivative with respect to $t$ for everything after the $\frac d{dt}$:

$$\frac d{dt}x^2$$

But if there is something in the numerator, like

$$\frac{dx}{dt}x^2$$

It means to take the derivative of the numerator with respect to $t$, then multiply it by whatever comes after.

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    Thanks alot, could you please further specify what is the difference between $\frac{dx}{dt}x^2+x$ and $\frac{d}{dt}x^2+x$ i don't quite understand what it means to take the derivative of the numerator2017-02-22
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    @simply beautiful art. (+1) I was taught integration was the reverse of differentiation and sadly the FTC didn't mean anything to me. Integral and differnetial calculus should be kept separate as you allude to in my opinion2017-02-22
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    @TheNotoriousSc the first says to take the derivative of $x$, then multiply by $x^2$ and add $x$. The second says to take the derivative of $x^2+x$.2017-02-22
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In all, I saw $5$ mathematical expressions:

  1. The first one is a definite integral of $x^2$ from $a$ to $b$.
  2. No meaning as it contains no differential. If it contained a differential, it would be an indefinite integral or an antiderivative.
  3. No meaning due to above reason.
  4. Derivative of $x^2$ with respect to $t$
  5. Product of $x^2$ with the derivative of $x$ with respect to $t$

These are what they actually are.

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The users above have sufficiently explained the difference between the integral notations, so I won't go over it again, but you asked for an explanation of the difference between $\frac {dx}{dt} x^2$ and $\frac d{dt} x^2$.

Well let's say that $x = f(t) = 2t^2$. Then the evaluation of the first expression would be to differentiate $x$ with respect to $t$ and multiply it by $x^2$: $\frac{dx}{dt} \times x^2 =4tx^2$.

The second one is a bit harder to evaluate. In this one, we are differentiating $x^2$ with respect to $t$. However, $x$ is a function of $t$ and does not contain any terms of $t$ in it, so how do we differentiate it? Well, in this case, we can rewrite $x^2$ using the function of $t$, but in other cases, we can't just do that. Instead, we use a method called implicit differentiation.

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    Is $\frac d{dt} x^2$ equivelent with $\frac {dx^2}{dt}$2017-02-22
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    @TheNotoriousSc As far as I know, yes.2017-02-22
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    @TheNotoriousSc Yes, but we would never write the second notation. We only use the second notation when we are differentiating a function of a variable, say f(_x_), with respect to that variable. Otherwise, the first notation would be used.2017-02-22