Question

How do you evaluate `d /dx \int_[x]^{x^4} sqrt{t^2+t} dt`?

Answer 1

To solve `d /dx \int_x^{x^4} sqrt{t^2+t} dt`, we will use the Fundamental Theorem of Calculus.

`d /dx \int_x^{x^4} sqrt{t^2+t} dt`
`= [\sqrt{(x^4)^2+x^4}] * (4x^3) - [\sqrt{(x)^2+x}] * (1)`
**don't forget to chain!!

Answer 2

`d/dx \ int_x^(x^4) \ sqrt(t^2+t) \ dt = 4x^3sqrt(x^8+x^4) - sqrt(x^2+x)`

Explanation:

If asked to find the derivative of an integral using the fundamental theorem of Calculus, you should not evaluate the integral.

The Fundamental Theorem of Calculus tells us that:

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

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

We are asked to find:

`d/dx \ int_x^(x^4) \ sqrt(t^2+t) \ dt`

(notice the upper and lower bounds are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral as follows:

`int_x^(x^4) \ sqrt(t^2+t) \ dt = int_0^(x^4) \ sqrt(t^2+t) - int_0^(x) \ sqrt(t^2+t) \ dt`

We have arbitrary chosen the lower limit as `0` wlog (any number will do!). The second integral is is now in the correct form, and we can directly apply the FTOC and write the derivative as:

`d/dx \ int_0^(x) \ sqrt(t^2+t) \ dt = sqrt(x^2+x)`

And using the chain rule we can write:

`d/dx int_0^(x^4) \ sqrt(t^2+t) = (d(x^4))/dx d/(d(x^4)) int_0^(x^4) \ sqrt(t^2+t)`

Now, `(d(x^4))/dx=4x^3`, And, using the FTOC, we have:

`d/(d(x^4)) int_0^(x^4) \ sqrt(t^2+t) = sqrt((x^4)^2+(x^4))`

Hence combining these trivial results we get:

`d/dx \ int_x^(x^4) \ sqrt(t^2+t) \ dt = 4x^3sqrt(x^8+x^4) - sqrt(x^2+x)`