Let F(x)=∫a^u(x) f(t) d t for the specified a, u, and f . Use a CAS to perform the following ste

Let F(x)=∫a^u(x) f(t) d t for the specified a, u, and f ....

Key Concepts

Fundamental Theorem of Calculus

This theorem connects differentiation and integration. When a function is defined as an integral with a variable upper (or lower) limit, the theorem allows us to differentiate the integral by evaluating the integrand at that variable limit. It is the foundation for computing derivatives of functions defined by integrals.

Chain Rule

The chain rule is used in differentiation when the function being differentiated is composed with another function. In the context of an integral with a variable limit, after applying the Fundamental Theorem of Calculus, the derivative of the limit function must be multiplied by the derivative of the upper limit. This combination is key to finding the first derivative of the composite integral function.

Domain of Functions

Determining the domain involves identifying all x values for which the function is defined. For functions defined by integrals, this means ensuring that the limits of integration (including any variable ones) lie within the domain of the integrand. In cases where the integrand is a polynomial, the domain may extend over all real numbers, but careful attention must be paid to the structure of the limit functions.

Critical Points and Increasing/Decreasing Intervals

Critical points occur where the first derivative is zero or undefined. After computing the derivative using the Fundamental Theorem of Calculus and the Chain Rule, analyzing the sign of the derivative reveals where the function is increasing or decreasing. This analysis is essential for understanding the local extrema and the overall behavior of the function.

Concavity and Points of Inflection

The second derivative of a function provides information about its concavity. By computing the second derivative and determining where it is zero or changes sign, one can identify inflection points and classify intervals as concave up or concave down. This information helps in sketching the graph of the function and understanding its curvature.

Graphical Interpretation of Derivative Information

Interpreting results from the first and second derivatives guides the sketching of the function’s graph. The locations of increasing/decreasing behavior, local extrema, and points of inflection, together with domain constraints, provide a comprehensive picture of the function’s behavior. Plotting based on these factors offers a visual validation of the analytical results.

Transcript

00:02 So in this problem, we are given this function big of x to be equal to the integral from a constant zero to a function u of x of a function f ft with respect to t, and we are told that a is zero, u of x is 1 minus x squared, and f f t is going to be this polynomial function.

00:26 So the first part here in this problem is to find the domain of f.

00:34 Later on we're going to learn more about big f of x.

00:41 So first thing is to find the domain.

00:44 Since this is all a polynomial and so we can solve for f here, but by integrating, but since we know that everything here is a polynomial, we can say that the domain of f is from minus infinity to infinity.

01:24 So next part, they ask us to calculate f prime of x and to determine zeroes.

01:32 So first thing we're going to do, f prime is.

01:37 Is derivative with respect to x of big f of x.

01:45 So what that means, now that we have a definition for f of x, that is this, that is this, so we have the integral from zero to one minus x squared of f f f x evaluated at t that is t squared minus two times t minus three with respect to t so for this part we are going to recall a definition of integral and that is not here but but if we have a direct of an integral from a constant to a function of x of f of t so that is the setup right here that is going to be equal to let's write it out over here so we have a derivative with respect to x the constant a of 2 a function u of x a function f of t with respect to this is going to be equal to to f evaluated at u of x and multiplied by derivative with respect to x of u of x so since this is the setup that we have this is what we are going to substitute over there over here so so it is this function of ft evaluated at ufx so let's start with t squared now instead of t we have one minus x squared n squared that's this squared minus two times instead of t t is now 1 minus x squared and minus 3.

04:37 Alright, now we're multiplying the derivative with respect to x of the function u, and u is 1 minus x squared.

04:52 Alright, so let's work on expanding this part.

04:58 So here we have a binomial squared.

05:01 So we begin with first term squared plus first time first time times the second square minus two x squared and the second plus the second term squared since it's x squared that is to the fourth power now that is all this first term now.

05:39 Now for this first term now for this one we are going to distribute this minus two so we have minus two times one minus two times minus x squared we have plus two x squared and the constant term we don't do anything there all right so we expanded this part and now we can find the derivative here derivative of the first term of zero and the second one it is a power of x so we bring down this two this negative is from here and now instead of x squared it's just x all right so here i am going to this 2x minus 2x squared cancels with this plus 2x squared and we have several constants and then just x raised to the fourth power so we have x to the fourth power one minus two minus three that is minus four and our second factor minus two times x all right so then this equal we can substitute um not substanti sorry, we are going to distribute and then we are left with 8x, 8 times x minus 2 times x to the 5th power.

07:37 So this is our function that we're looking for, that's f prime.

07:46 Now to 2 minute zeros, i want to graph it.

07:52 So this is this function in red is, is, is f prime and its zeros are here, here, and here.

08:04 Those are the calculator tells me that they are negative 1 .4140 and positive 1 414.

08:24 Now the other thing that this part asks us is which part in its domain is f increasing and decreasing.

08:33 So, when f prime is zero, it is neither increasing or decreasing.

08:45 But in this, from negative infinity to 1 .14, f prime is positive.

08:59 So that means f is increasing.

09:02 Now for this part over here, that's up to here.

09:11 F prime is negative, so f is increasing.

09:21 And from 0 to positive, 1 .44, this f prime is positive...