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Μάθημα 8: Properties of definite integrals- Negative definite integrals
- Finding definite integrals using area formulas
- Finding definite integrals using area formulas
- Definite integral over a single point
- Integrating scaled version of function
- Switching bounds of definite integral
- Integrating sums of functions
- Worked examples: Finding definite integrals using algebraic properties
- Εύρεση ορισμένου ολοκληρώματος χρησιμοποιώντας αλγεβρικές εκφράσεις
- Definite integrals on adjacent intervals
- Worked example: Breaking up the integral's interval
- Worked example: Merging definite integrals over adjacent intervals
- Ορισμένα ολοκληρώματα σε διπλανά διαστήματα
- Functions defined by integrals: switched interval
- Finding derivative with fundamental theorem of calculus: x is on lower bound
- Finding derivative with fundamental theorem of calculus: x is on both bounds
- Functions defined by integrals: challenge problem
- Επανεξέταση ιδιοτήτων ορισμένων ολοκληρωμάτων
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Integrating scaled version of function
Sal uses a graph to explain why we can take a constant out of a definite integral.
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- [Voiceover] We've already
seen and you're probably getting tired of me
pointing it out repeatedly, that this yellow area right
over here, this area under the curve y is equal
to f of x and above the positive x-axis or I
guess I can say just above the x-axis between x
equals a and x equals b, that we can denote this area
right over here as the definite integral of from a to b of f of x dx. Now what I want to explore
in this video and it'll come up with kind of an
answer that you probably could have guessed on
your own, but at least get an intuition for it,
is that I want to start thinking about the area under the curve that's a scaled version of f of x. Let's say it's y is
equal to c times f of x. Y is equal to some number times f of x, so it's scaling f of x. And so I want this to be
kind of some arbitrary number, but just to help me
visualize, you know I have to draw something so I'm
just gonna kind of in my head let's just pretend the c is a three for visualization purposes. So it's going to be three
times, so instead of one, instead of this far right over here it's going to be about this far. For right over here, instead
of this far right over here it's going to be that
and another right over there. And then instead of it's
going to be about there. And then instead of it
being like that it's going to be one, two and then
three, right around there. So I'm starting to get a
sense of what this curve is going to look like, a
scaled version of f of x. And at least what I'm
drawing is pretty close to three times f of x, but
just to give you an idea is going to look something like, and let's see over here if this distance, do a second one, a third
one, is gonna be up here. It's gonna look something like this. It's gonna look something like that. So this is a scaled version
and the scale I did right here I assumed a positive c greater than zero, but this is just for
visualization purposes. Now what do we think the area under this curve is going to be between a and b? So what do we think this area right over here is going to be? Now we already know how we can denote it. That area right over there
is equal to the definite integral from a to b of the function we're integrating is c f of x dx. I guess to make the question
a little bit clearer, how does this relate to this? How does this green area
relate to this yellow area? Well one way to think
about it is we just scaled the vertical dimension up by
c, so one way that you could reason it is if I'm finding
the area of something, if I have the area of a
rectangle and I have the vertical dimension is
let's say I don't want to use those same letters
over and over again. Well let's say the
vertical dimension is alpha and the horizontal dimension is beta. We know that the area is
going to be alpha times beta. Now if I scale up the
vertical dimension by c, so instead of alpha this
is c times alpha and this is, the width is beta,
if I scale up the vertical dimension by c so this
is now c times alpha, what's the area going to be? Well it's going to be c alpha times beta, or another way to think of it, when I scale one of the
dimensions by c I take my old area and I scale
up my old area up by c. And that's what we're doing, we're scaling up the vertical dimension by c. When you multiply c times f of x, f of x is giving us the vertical height. Now obviously that
changes as our x changes, but when you think back
to the Reimann sums the f of x was what gave us
the height of our rectangles. We're now scaling up the
height or scaling I should say because we might be scaling
down depending on the c. We're scaling it, we're
scaling one dimension by c. If you scale one dimension by c you're gonna scale the area by c. So this right over here, the integral, let me just rewrite it. The integral from a to b of c f of x dx, that's just going to be
the scaled, we're just going to take the area of f of x, so let me do that in the same color. We're going to take the
area under the curve f of x from a to b f of x dx and we're just going to scale it up by this c. So you might say, "Okay
maybe I could have felt "that was, you know, if I
have a c inside the integral "now I can take the c
out of the integral", and once again this is
not a rigorous proof based on the definition of
the definite integral, but it hopefully gives you a little bit of intuition why you can do this. If you scale up the
function, you're essentially scaling up the vertical
dimension, so the area under this is going to just be a scaled up version of the area under
the original function f of x. And once again really,
really, really useful property of definite
integrals that's going to help us solve a bunch
of definite integrals. And kind of clarify what
we're even doing with them.