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# Let F (x) = integral from 0 to x^(1/3) of function sq root (1 + t^4) dt. F'(0) = ?

Let F (x) = integral from 0 to x^(1/3) of function sq root (1 + t^4) dt. F'(0) = ?

Solution manual says: F (x) = f(g(x)), where g(x) = x^(1/3) and f (z) = integral from 0 to z of function sq root (1+t^4) dt.  Then they apply chain rule.  Do not understand

### 1 Answer by Expert Tutors

Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
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We are given the function

F(x) = int( sqrt(1+t4), t=0..x1/3)

We do not need to evaluate this integral, since we only want the derivative F'(x) and can therefore use the Fundamental Theorem of Calculus.

Use the substitution z=x1/3, then

F(z(x)) = int( sqrt(1+t4), t=0..z)

By the chain rule,
F'(x) = dF/dx = dF/dz * dz/dx

We have

dz/dx = (1/3)x-2/3

and, by the Fundamental Theorem,

dF/dz = sqrt(1+z4) = sqrt (1+x4/3)

so that

F'(x) = sqrt (1+x4/3) * (1/3)x-2/3

Because of the term x-2/3,  F'(0) is undefined.

Thank you Andre.  A couple of follow up questions as I am studying for an actuarial certification and my calculus is rusty.

1. Why do we need to substitute z=x^1/3?

2.  Towards the end, dF/dz = sqrt (1+z^4) = sqrt (1+x^4/3)  Confused on this step.  What happens to the t variable?
Those are excellent questions, Rob. They both concern the nature of the Fundamental Theorem. One thing you may remember from calculus is that derivative and integral are in some sense inverse operations: the integral
of the derivative of a function is just that function (up to an additive constant) and the derivative of an integral of a function is that function. For indefinite integrals (antiderivatives) this statement is true by definition. For definite integrals, usually interpreted as “area under the graph” the statement is only true if we turn the integral into a function, by letting its upper limit be a variable, say z: then F(z) = int (f(t), t=0..z) is a new function, defined for all z where the integral exists.

Notice that the integration variable t here is only internal to the integral; the function F(z) does not depend on it. We could change the letter of the integration variable (e.g., from t to u) without changing the function F(z). This is why an integration variable is often referred to as a "dummy variable."

The Fundamental Theorem, in one of its many formulations, now says that F(z) is differentiable and

F’(z) = f(z)

To apply the theorem, wherever you see the integration variable t in the integrand, you need to change it to the free variable z. This eliminates t from the problem (answering your last question first).

Now what if the upper limit of the integral is not just z, but some more complicated function of some other variable x? The standard way of dealing with this is to define z to be this more complicated function, z=z(x).
This means that the integral is now a function of z, which itself is a function of x, F=F(z(x)). In cases where you have a composition of two functions, you need to use the chain rule to find the derivative with respect to x. Therefore,
F’(x) = dF/dz * dz/dx.

In the end, everything needs to be expressed in terms of x, which was our original variable, so in the dF/dz part I had to substitute for z the definition in terms of x.
I hope this makes sense. Please ask if you have more questions!
Rob,

Recall the Fundamental Theorem of Integral Calculus states that if f(x) is defined and continuous on a closed interval [a,b], and if F(x) is defined on this interval by

F(x) = ∫0x f(t) dt,

Oh my.  We are like the Three Mathsketeers over here!  All comments for one.
-HH-
Thank you again Andre and thank you Kirill and Hassan as well.  This helped a lot!  I will continue to work through this and will let you know if I have an further questions