(a) Use integration by parts to show that ∫f(x) d x=x f(x)-∫x f^'(x) d x (b) If f and g are inve

(a) Use integration by parts to show that ∫f(x) d x=x...

Key Concepts

Integration by Parts

Integration by parts is an essential method in calculus used to integrate products of functions. It is based on the product rule for differentiation and allows the transformation of a complex integral into a simpler one by choosing appropriate parts of the integrand to differentiate and integrate. This technique is particularly useful for functions where direct integration is challenging.

Inverse Functions

Inverse functions reverse the effect of the original function, swapping the roles of the dependent and independent variables. In the context of integration, properties of inverse functions, such as the relationship between their derivatives, enable the derivation of formulas that connect the integrals of a function and its inverse. This relationship leverages the symmetry between a function and its inverse, which is a key insight in many calculus problems.

Substitution Method in Integration

The substitution method involves changing the variable of integration to simplify an integral, often by aligning the integrand with a known form. When dealing with inverse functions, substituting the original function with a new variable can reveal underlying structures and symmetries in the integral, making it possible to transform and evaluate complex integrals with greater ease.

Geometric Interpretation of Integral Relationships

Geometric interpretation in calculus provides visual insights into the relationships between functions and their integrals. For inverse functions, reflecting the graph across the line y = x can illustrate how areas under the curve of one function correspond to areas under the curve of its inverse. This visualization helps to understand the balance and symmetry inherent in the integral formulas, offering a tangible explanation of abstract integration relationships.

Transcript

00:01 So we're going to be used integration with parts to kind of show a whole lot of things so we can get a better understanding of integration by parts so first i want to show that the integral of ffx dx will equal this function remember the integration of my parts formula is written out in green in case you forgot the angle of ffx dx equals uv minus the angle of v d u so if i lit from my original function f of x to equal u and my dv is going to equal you and my dv is going to equal dx so if i multiply my u and dv that's still f of x dx so then i'll get f prime of x d x will equal d u and my v will equal x so i differentiated my u and i integrated my vs so i can rewrite my formula f of x dx using the integration of my parts so my uv or my yeah you my uv so f of x times x minus the integral of v to u so x f prime of x d x and i can rewrite that to say x times f of x minus that integral so we got what we want to show i'm going to call this part equation one because i may use this in the future and that's i'm just going to call this part a since that's the first thing i want to show so let's show in part b that i can if we put some bounds on this integral well this is just going to be times f of b minus a times f of a minus the integral from f of a to f of b g y d y d y all right so i'm using equation one we are in the form we want so let's just put this in bound so this is x f of x this is all evaluated from a to b minus the integral from a to b of x minus the integral from a to b of x of x f prime of x d x so looking more over the x f of x if we just evaluate that how from we're using calculus this is going to be b f of b minus a f of a minus still this angle from a to b f x f prime of x d x d x so this side we have matches what we want so now we just got to work on the interval so i'm going to let my dependent variable y this is going to equal f of x so now d y over d x is only equal f prime of x and then how i can just say well d y is just going to be f prime of x d x so i just took i let my variable y equals the function f of x took the derivative and solve for d y so assuming that this function is one to one we can take the inverse of the function so if we take the inverse of the function so if we take the inverse of the function the function we'll get that x will equal f inverse of y which we can rewrite as saying g of y so now when we let x equal a and when we get so when x is a we get that y is f of a and when we let x b b b we get we get y as f of b so we can rewrite our entire equation to say that this is b f of a minus a f of b minus the integral from f of a times f of b g of y d y d y all right let me just call this equation two and i guess a little caveat should put it is where x equals f inverse of y which is equal to g of y so part c is i want to take a more geometric approach of what this integration of our parts looks like so i'm going to quickly pause the video all right so this is a nice little picture of how the integration of my parts is working so we have this little square and we can find that the area of the square or the rectangle but this is going to be b times f of b well this area of the square is also going to equal the sum of these three little sections are my a that called a b and c so i know b times f of b well that's going to equal a plus b plus c all right so now let's try to figure what each one of these are so a well this is just going to be the integral i guess i should call this line f of x so this line here from a to b is f of x so that's going to be the angle from a to b of f of x d x the angle b if we think of this as integrating with respect to y, this is going to be f of a to f of b g y d y and the last one the area of c is again is just an xl square so this should be a times f of a so if we substitute in what we know what we have back into this equation we got b f of b this is going to equal the integral from a to b f of x dx plus the integral from f of a to f of b g y d y plus a times f of a all right so if we subtract a f of a plus this integral f of a f of b g y d y to both sides we will create a new equation so this is b f of a minus the enter a f of b, or that is b f of b minus a f of a minus the integral from f of a to f of b, g y, gy, dy, which this is the same as the equation two that we got earlier in part b.

07:55 So these are the same equations.

08:00 All right.

08:02 So let's call it last part part d...

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