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If we want to take the derivative $d/dx$ of, say, $\int_{0}^{1} x^2dx$, what will the answer be? does that mean to take $g(x) = \int_{0}^{1}x^2dx$ and find $g'(x)$? If so, we have $g(t) = \int_{0}^{1}t^2dx = t^2$ and therefore the derivative $d/dx$ of $\int_{0}^{1} x^2dx$ is $2x?$
What needs to be done first, calculate the integral to get $g(x) = 1/3$ or do what I did and why? thanks

Omer
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    That integral is a numerical constant, not a function of $x$, so yes. It's derivative is $0$. This is true for any definite integral. – lulu Jun 09 '19 at 12:58

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$$\int_{0}^{1} x^2dx $$ is just a number - it happens to be $1/3$. It does not depend on $x$ and its derivative (should you think of it as a constant function) is $0$.

I think you are confusing this definite integral with the indefinite integral

$$ f(x) = \int_{0}^{x} t^2dt . $$ The derivative of that function is $x^2$, which you know from the fundamental theorem of calculus even without guessing the antiderivative $x^3/3$.

Edit in response to a comment: see this answer about "dummy variables" What exactly is a 'dummy variable'? .

Ethan Bolker
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  • I know, If I wrote $g(t) = \int_{0}^{1} x^2dx$ it was OK for me, but what annoys me is when I write $g(x) = \int_{0}^{1}x^2dx$ why doesn't it make it $g(1) = \int_{0}^{1}1^2dx$? – Omer Jun 09 '19 at 13:01
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    The $x$ in the integral is just a dummy variable of integration (i.e. you could replace the integration variable with any symbol and it would be the same, e.g. $\int_0^1 x^2, dx = \int_0^1 u^2, du = \int_0^1 y^2, dy = \int_0^1 \Upsilon^2, d\Upsilon = \int_0^1 \ddot{\smile}^2, d\ddot{\smile}=\cdots$). It is separate and different in meaning to the $x$ in your $g$ function (and it's nicer really here to use a different symbol for the $g$ input and the integration variable). – Minus One-Twelfth Jun 09 '19 at 13:01
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  • A definite integral like $\int_a^bf(x)\,dx$ is a number; the dummy variable one uses does not affect the value of the definite integral: $$ \int_0^1x^2\,dx=\int_0^1y^2\,dy=\int_0^1t^2\,dt=\cdots. $$

  • Given any real constant $C$, one can define a constant function $$ g(x)=C\quad \textrm{for all the real number $x$}. $$ The derivative of a constant function (at any point) is $0$.

  • Writing $g(x)=\int_0^1x^2\,dx$ is bad; you are using the symbol $x$ for two different kinds of purposes: on the right hand side, it is a dummy variable; on the left, it is used as a variable of the function $g$. If such expression makes sense at all, it means the same thing as $g(x)=\int_0^1u^2\,du$, which defines a constant function, the derivative (at any point) of which is $0$.

  • The definite integral $\int_0^1x^2\,dx$ and $\int_0^1t^2\,dx$ are two different things: the value of the later depends on the value of $t$. In fact, $$ \int_0^1x^2\,dx=\frac13,\quad\int_0^1t^2\,dx=t^2. $$

  • "Taking the derivative" and applying the differential operator $d/dx$ to a function are not quite the same. One can take the derivative of any differentiable function but one cannot applying the differential operator $d/dx$ to a function that does not have $x$ as its variable. For instance, it does not make sense to, say, take $d/dx$ to $g(t)=t^2$. But it make sense to write $g'(t)=2t$.

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There appears to be a major point of confusion here. The integral $$ \frac{1}{3} = \int_0^1 x^2 dx$$ is a number. It's exactly equal to $1/3$. Taking the derivative of a constant is zero (since the constant function does not change).

Writing notation like $$ g(x) = \int_0^1 x^2 dx$$ is very poor notation, since the $x$ appearing in the integral is not the same variable as appears in $g(x)$.

davidlowryduda
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