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### Course: Calculus 2 > Unit 1

Lesson 6: Fundamental theorem of calculus and accumulation functions- The fundamental theorem of calculus and accumulation functions
- Functions defined by definite integrals (accumulation functions)
- Functions defined by definite integrals (accumulation functions)
- Finding derivative with fundamental theorem of calculus
- Finding derivative with fundamental theorem of calculus
- Finding derivative with fundamental theorem of calculus: chain rule
- Finding derivative with fundamental theorem of calculus: chain rule

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# Functions defined by definite integrals (accumulation functions)

Understanding that a function can be defined using a definite integral. Thinking about how to evaluate functions defined this way.

## Want to join the conversation?

- why do we end with, let's say, dx?(6 votes)
- Imagine a curve f(x) with a series of rectangles underneath it. The height of the rectangle would be f(x) and let us call the width of the rectangle delta x. The thinner you make your rectangles, the more rectangles you can fit within the area under the curve. Now, lets assume that each rectangle had an infinitesimally thin width, which we will call dx. Now, you get a perfect answer to the area under the curve. So the area of each rectangle is equal to f(x) dx (where dx is the width of the super thin rectangles). Now to find the total area under the curve, you have to add up all the areas of the rectangles which is denoted by ∫f(x) dx. Hope this helps! (Note: that ∫ is essentially an elongated "S" which stands for sum).(31 votes)

- Wouldn't the areas cancel out since one side is negative and the other side is positive?(6 votes)
- Nope, because even when t is -ve the area under the graph is in +ve y-axis so th areas would simply add up.(5 votes)

- What is g(-3). Is it a positive or a negative number?(4 votes)
- https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-6/v/switching-integral-bounds

If the top bound is less than the bottom bound then yeah, it's the negative of if the bounds were more standard.(4 votes)

- If the x-axis and the t-axis is the same axis, aren't "t" and "x" the same thing? Why don't we just use one or the other in the integral expression?(5 votes)
- why 21? the height is 4 and the width is 5, so I believe the integral should be 20.(0 votes)
- You're forgetting about the little triangle-shaped section on top of the rectangle.(19 votes)

- When the graph is under x axis will that value be negative?(3 votes)
- Yes. In the instances you want it to be positive you have to find each spot it crosses the x axis as solve each section seperately(2 votes)

- Why was -2, at2:41, used as the lower boundary in the definite integral to define g(1)? It seems quite arbitrary.(2 votes)
- Yes, because it is just an example of how you can do it. It could also be 0 or 1 or even a function of
`x`

.(3 votes)

- Why is it that the definite integral from a to b of f(t) with respect to t is always the area underneath the curve from t=a to t=b?(2 votes)
- If you haven't watched the parts on riemann sums thse explain it. But basically the riemann sums split the area underneathe a graph into smaller and smaller rectangles, until they fit the graph exactly and thus give the exact area. the summation formula that is used is then used for integrals.(3 votes)

- If a graph, is just a visual representation of a function, then what actually is the area under the curve if we don't draw it out. In this problem, we only know the area under the curve because f(t) was drawn out. How could we figure this out without drawing the function out?

The area under the curve represents something, but I forgot, if we find out what that is then that is equal to the area under the curve.

I think it is the value of the function of the antiderivative at that same x value.

That makes sense why the integral is equal to the area under the curve them. The integral is just the antiderivative. The integral of f(x) is equal to the antiderivative of f(x) which is written as F(x).(2 votes)- Integration starts to get really helpful in things like physics. If you are learning basic physics right now, you would know that the area under the graph of a v-t graph is equal to the displacement. the graph of something like a velocity-time graph would be really simple. But, in real life this is extremely untrue. When driving a car, the acceleration is not constant throughout time. If one can represent this as a function and solve for its integral, they can easily figure out the displacement of a moving body.(1 vote)

- If the curve lies below the x-axis do we still count it, or do we only calculate the area above the x-axis?(2 votes)

## Video transcript

- [Instructor] You've
already spent a lot of your mathematical lives
talking about functions. The basic idea is give a
valid input into a function, so a member of that function's domain, and then the function is going
to tell you for that input what is going to be the
corresponding output. And we call that
corresponding output f of x. So, for example, there's many
ways of defining functions. You could say something like
f of x is equal to x squared. So that means that whatever x, whatever you input into the function, the output is going to
be that input squared. You could have something
defined like this. F of x is equal to x squared if x odd. And you could say it's equal
to x to the third otherwise, otherwise. So if it's an odd integer, it's an odd integer, you just square it. But otherwise, for any other real number, you take it to the third power. This is a valid way of
defining a function. What we're going to do in this
video is explore a new way or potentially a new way for
you of defining a function. And that's by using a definite integral, but it's the same general idea. So what we have graphed
here, this is the t-axis, this is the y-axis, and we have
the graph of the function f, or you could view this as the graph of y is equal to f of t. Now, what I want to, and this is another way of representing what outputs you might
get for a given input. Here, if t is one, f of t is five. If t is four, f of t is three. But I'm now going to define a new function based on a definite integral of f of t. Let's define our new function. Let's say g, let's call it g of x. Let's make it equal to
the definite integral from negative two to x of f of t dt. Now, pause this video,
really take a look at it. This might look really fancy,
but what's happening here is, given an input x, g of x
is going to be based on what the definite integral
here would be for that x. And so we can set up a little table here to think about some potential values. So let's say x, and let's
say g of x right over here. So if x is one, what is g of x going to be equal to? All right, so g of one is going to be equal to
the definite integral, going from negative two. Now x is going to be equal
to one in this situation. That's what we're inputting
into the function. So one is our upper bound of f of t dt. And what is that equal to? Well, that's going to be the area under the curve and above the t-axis, between t equals negative
two and t is equal to one. So it's going to be this area here. And we, since it's on a grid, we can actually figure this out. We can actually break
this up into two sections. This rectangular section is
three wide and five high, so it has an area of 15 square units. And this little triangular section up here is two wide and one high. Two times one times one half, area of a triangle, this
is going to be another one. So that area is going to be equal to 16. What if x is equal to two? What is g of two going to be equal to? Pause this video, and
try to figure that out. Well, g of two is going to be
equal to the definite integral from negative two, and now
our upper bound's going to be our input into the function
to two, of f of t dt. So that's going to be going from here, all the way now to here. And so it's the area we just calculated. It's all of this stuff, which we figured out was 16 square units, plus another one, two, three,
four, five square units. So 16 plus five, this is
going to be equal to 21. So hopefully that helps, and the key thing to appreciate
here is that we can define valid functions by using
definite integrals.