Question

How do you calculate the derivative of `int(2t-1)^3 dt` from `t=x^2` to `t=x^7`?

Answer 1

`(2x^7 - 1)^3 - (2x^2 - 1)^3`

Explanation:

Let `f(t) = (2t - 1)^3`. Then, if `F(t)` an antiderivative of `f(t)`, it holds that:

`F'(t) = f(t)`.

We also know that `int f(t)dt = F(t) + c`. But also,

`(F(t) + c)' = f(t)` because `(c)' = 0` for any `cinRR`.

Therefore,

`[int f(t)dt]' = f(t)`,

so evaluating the derivative of that integral is exactly the same as evaluating `f(t)` at the two given `t` points, then taking their difference (as the question says to evaluate from `x^2` to `x^7`)

Finally,

`{[int f(t)dt]'}_(x^2)^(x^7) = f(x^7) - f(x^2)`

`=(2x^7 - 1)^3 - (2x^2 - 1)^3`

Note that further simplification is applicable (but not required) since this is a difference of cubes.

Answer 2

`= x( 7x^5 (2x^7-1)^3 - 2 (2x^2-1)^3 )`

Explanation:

`d/dx ( int_(x^2)^(x^7) (2t-1)^3 dt )`

We can use the Fundamental Theorem of Calculus , Part 1:

`d/dx ( int_(a)^(x) f(t) \ dt ) = f(x)`

But if it is:

`d/dx ( int_(a)^(u) f(t) \ dt )` where `u = u(x)`, then we must also use the chain rule:

`d/dx ( int_(a)^(u(x)) f(t) \ dt ) = d/(du) ( int_(a)^(u) f(t) \ dt ) (du)/dx = f(u(x)) cdot (du)/dx`

To make this particular integral fit within the FTC, we need also do some interval manipulation:

`d/dx ( int_(x^2)^(x^7) (2t-1)^3 dt )`

Splitting out the integration intervals:

`= d/dx ( int_(x^2)^(0) (2t-1)^3 dt + int_(0)^(x^7) (2t-1)^3 dt )`

Switching the first interval limits:

`= - d/dx ( int_0^(x^2) (2t-1)^3 dt ) + d/dx ( int_(0)^(x^7) (2t-1)^3 dt )`

Applying the FTC:

`= - (2(x^2)-1)^3 d/dx(x^2) + (2(x^7)-1)^3 d/dx (x^7)`

`= (2x^7-1)^3 cdot 7 x^6 - (2x^2-1)^3 cdot 2 x`

`= x( 7x^5 (2x^7-1)^3 - 2 (2x^2-1)^3 )`

Maybe that can be simplified a bit more.