Question

What is the derivative of `int_x^(x^2) t^3 \ dt` wrt `x`?

Answer 1

`d/(dx) int_x^(x^2) t^3 dt =x^6-x^3`

Explanation:

According to the `2^(nd)` Fundamental Theorem of Calculus,
`d/(dx) int_a^b f(t) dt = f(b) - f(a)`.

Using your example, `d/(dx) int_x^(x^2) t^3 dt =(x^2)^3 - x^3 = x^6 - x^3`.

Answer 2

`d/dx \ int_x^(x^2) t^3 \ dt =2x^7 - x^3`

Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

`d/dx \ int_a^x \ f(t) \ dt = f(x)` for any constant `a`

(ie the derivative of an integral gives us the original function back).

We are asked to find:

`d/dx \ int_x^(x^2) t^3 \ dt`

Initially we can manipulate the integral as follows (although we have chosen `0` as the lower limit we could in fact choose any constant:

`int_0^(x^2) t^3 \ dt = int_0^x t^3 \ dt + int_x^(x^2) t^3 \ dt`

And so:

`int_x^(x^2) t^3 \ dt = int_0^(x^2) t^3 \ dt - int_0^x t^3 \ dt`

And therefore differentiating we get:

`d/dx \ int_x^(x^2) t^3 \ dt = d/dx \ int_0^(x^2) t^3 \ dt - d/dx \ int_0^x t^3 \ dt`

(notice the upper bounds of the first integral are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral using a substitution and the chain rule. Let:

`u=x^2 => (du)/dx = 2x`

The substituting into the first integral we get:

`d/dx \ int_x^(x^2) t^3 \ dt = d/dx \ int_0^u t^3 \ dt - d/dx \ int_0^x t^3 \ dt`

`" " = (du)/dx*d/(du) \ int_0^u t^3 \ dt - d/dx \ int_0^x t^3 \ dt`

`" " = 2xd/(du) \ int_0^u t^3 \ dt - d/dx \ int_0^x t^3 \ dt`

And now the derivative of the integral is in the correct form for the FTOC to be applied, giving:

`d/dx \ int_x^(x^2) t^3 \ dt = 2x \ u^3-x^3`
`" " = 2x \ (x^2)^3-x^3`
`" " = 2x \ x^6 - x^3`
`" " = 2x^7 - x^3`

Answer 3

Another way to do this is to apply Leibniz's integral rule:

`d/(dx)[int_(a(x))^(b(x)) f(x,t)dt] = int_(a(x))^(b(x)) del/(delx)[f(x,t)]dt + f(x,b(x))(db)/(dx) - f(x,a(x))(da)/(dx)`

In this case, we have:

`d/(dx)[int_(x)^(x^2) t^3 dt]`,

where `a(x) = x`, `b(x) = x^2`, and `f(x,t) = t^3`.

We obtain, by plugging in `a(x)` and `b(x)` in place of `t` for the `f(x,a(x))` and `f(x,b(x))` terms:

`= cancel(int_(x)^(x^2) 0dt)^(0) + stackrel(|[t^3]|_(t = x^2))overbrace((x^2)^3)cdotcolor(red)((db)/(dx)) - stackrel(|[t^3]|_(t = x))overbrace((x)^3)cdotcolor(red)((da)/(dx))`

since the derivative of `t^3` with respect to `x` is `0` (`t` does not vary when `x` varies), and the integral of `0` is `0` (a flat horizontal line along the `x` axis has no area).

We then take the derivative of `b(x) = x` with respect to `x` and `a(x) = x^2` with respect to `x` to get:

`=> color(blue)(d/(dx)[int_(x)^(x^2) t^3 dt]) = x^6 cdot color(red)(2x) - x^3 cdot color(red)(1)`

`= color(blue)(2x^7 - x^3)`