0

what is

$\int_0^x ax^2 dy$

additionally, how is this described when the variable of integration is not in the function? I'm not sure how to research this.

rocketboy2000
  • 39
  • 1
    To my knowledge, there's no real term for it, as you're treating $x$ as a constant while doing the integration. It's as if it were any other more familiar constant, like $\pi$ or $\sqrt 7$. – scoopfaze Jul 04 '19 at 17:43

4 Answers4

2

$$\int_0^x ax^2dy=ax^2\int_0^xdy=ax^2\cdot y\bigg|_0^x=ax^2(x-0)=ax^3$$

azif00
  • 20,792
1

$\int_{0}^{x} ax^{2} dy = a x^{2} \int_{0}^{x}dy = a x^{2}y|_{0}^{x} = ax^{3}$

Note that the function $f(x) = ax^{2}$ independent of the variable y, therefore, in this case, it is a constant function and follows the calculations above.

Allan Ramos
  • 249
  • 1
  • 18
1

In this expression $x$ is a free variable (that appears twice) and $y$ is a bound variable.

In general: $$\int_0^xcdy=[cy]^x_0=cx-c0=cx$$because the antiderivative of constant $c$ as a function on variable $y$ is $cy$.

Applying that here (where $c=ax^2$) we find that:$$\int_0^xax^2dy=ax^2x=ax^3$$

Further note that we can write also $\int_0^xcf(y)dy$ where function $f$ is prescribed by $y\mapsto1$ taking away the "objection" that the variable of integration is not present in the integrand.

So there is no essential need to pay any attention to that phenomenom.

drhab
  • 151,093
1

In terms of $y$, $ax^2$ is just a constant.

So without limits: $$\int ax^2 dy =ax^2y+C$$

Hence $$\int_0^x ax^2 dy =ax^2(x-0)=ax^3$$

Rhys Hughes
  • 12,842