# c2 integration to find area?Watch

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#1
how does finding the integral of a curve f(x) gives it's area between x coordinates 'a and b'?

it's not that i can't do the questions but i'm having trouble understanding why i am using intergration to find area under a curve, or rather how is it possible.

i believed the purpose of intergration was to go from f '(x) to f(x). that is why i can't imagine how intergration is being used to find area under f(x).

if someone could shed some light.
0
13 years ago
#2
well the integration (in simple terms) is found using the area... the area under a graph is divided into small rectangles of sides f(x) and dx.
the area under the graph is the sum of the areas of those small rectangles as dx is very small. so when u find the sum of those areas and take the limit as dx goes to 0, u will get the integral of f(x).
0
13 years ago
#3
i believed the purpose of intergration was to go from f '(x) to f(x). that is why i can't imagine how intergration is being used to find area under f(x).
that is one of the purposes, but when you put limits in, it can tell you the area under the curve - much more useful !
as yazan said, imagin the area is split up into lots of little bits using vertical lines (like you would do using the trapesium rule) and that all these bits are very thin rectangles.
The height of each rectangle is y, in this case f(x)
The width of each rectangle is a very small lengh along the x axis, so we say dx (d just means 'a little bit' really)
The area of a rectangle is height x width = f(x)dx
So if you add up the areas of all the rectangles, you get the area under the curve, = integral sign f(x)dx
and this is what you would recognise as integration
0
13 years ago
#4
(Original post by arslan)
how does finding the integral of a curve f(x) gives it's area between x coordinates 'a and b'?

it's not that i can't do the questions but i'm having trouble understanding why i am using intergration to find area under a curve, or rather how is it possible.

i believed the purpose of intergration was to go from f '(x) to f(x). that is why i can't imagine how intergration is being used to find area under f(x).

if someone could shed some light.

Consider a continuous function f(x) between the points f(a) and f(b), b>a. By the mean value theorem, there is a point c, a<c<b, such that f'(c)(b-a)=f(b)-f(a). If we integrate a function between two limits a and b then we are finding the limit of SUM(f(k(i))(x(i+1)-x(i))) as these partitions become infintesimal, where k(i) is a general point between x(i) and x(i+1). As we approach the limit, the number of value of k(i) that we can choose becomes small until we can only choose a single point that, by the MVT, must satisfy the condition above. The integral thus becomes (f(a+deltax)-f(a))+(f(a+2.deltax)-f(a+deltax))+...+(f(b)-f(b-delta.x))=f(b)-f(a).
0
13 years ago
#5
(Original post by Fly.)
Consider a continuous function f(x) between the points f(a) and f(b), b>a. By the mean value theorem, there is a point c, a<c<b, such that f'(c)(b-a)=f(b)-f(a). If we integrate a function between two limits a and b then we are finding the limit of SUM(f(k(i))(x(i+1)-x(i))) as these partitions become infintesimal, where k(i) is a general point between x(i) and x(i+1). As we approach the limit, the number of value of k(i) that we can choose becomes small until we can only choose a single point that, by the MVT, must satisfy the condition above. The integral thus becomes (f(a+deltax)-f(a))+(f(a+2.deltax)-f(a+deltax))+...+(f(b)-f(b-delta.x))=f(b)-f(a).
There's some mix-up there. I think by sum[f(k_i) (x_(i+1) - x_i)] you mean sum[f ' (k_i) (x_(i+1) - x_i)].
0
13 years ago
#6
I think you're just confusing yourself through familiarity with differentiation. The question, "why does integration give the area under the curve?" is similar to "why does differentiation give the gradient of a curve?" Do you know the answer to the second question? It's equally as confusing as the first, really, because all it is is some symbolic manipulation that magically gives you a useful result. That's all any mathematical operation is really.

You may have seen a proof (and if you haven't, it's worth a look) of why d/dx of x^n is nx^(n-1), and things like that. It uses delta-x's, and then takes limits as those deltas go to zero (i.e. the gradient between two points on the curve tends to the tangent as one point moves infintesimally close to the other). There is a similar proof for finding the area under a curve. It's basically the trapezium rule, but with infintesimally small widths of trapeziums, and the result is called "integration"! It's a standard thing you could probably find on the net, if "fly" hasn't already done it (it has the "mean value theorem", and i don't know what that is though!)

Key point is, it's not all that confusing once you recognise that your question isn't that far from something you're already familiar with.
0
13 years ago
#7
There's really no need to prove that integration gives the area under a curve, because integrals are defined the way they are (with partitions and rectangles that give better approximations to the curve as the partitions get smaller) to do just that. (At least in the sense of Riemann integration. )

What Fly. gave a proof of is that integrationis the reverse product of differentiation; that is, if f' is integrable on [a, b], then:
int(a,b) f'(x) dx = f(b) - f(a)

This is sometimes called the "second fundamental theorem of calculus", but it's really just some manipulation of the mean value theorem as the proof shows. And all that the mean value theorem says is if we have a function f that is differentiable on [a, b], then the tangent to the curve at some point in the interior of the interval is parallel to the secant line connecting f(a) and f(b).
0
13 years ago
#8
(Original post by arslan)
how does finding the integral of a curve f(x) gives it's area between x coordinates 'a and b'?
It just does.
0
13 years ago
#9
lol
0
13 years ago
#10
First thing to say is that a full justification of the connection between differentiation and integration is not trivial. I don't blame your teacher for glossing over it, mine did the same. If you go on to study this properly you'll understand why.

Initially differentition and integration were seen to be entirely different things. You say "i believed the purpose of intergration was to go from f '(x) to f(x)" - that's not true. Differentiation is the process of finding the rate of change of a function, while integration is the process of finding the area underneath a function. The connection between the two was a very important discovery, but it is not part of the definition of the operations. It turns out the be the case we can use derivitives to work out areas, and this is incredibly handy, but integration is not the "the reverse of differentiation."

Now I'll see if I can make this clear without getting too technical. You want to know why

The area under the curve f(x) between a and b is equal to G(b)-G(a), where G(x) differentiates to give f(x), that is G'(x) = f(x).

This follows from the following fact, called the Fundamental Theorem of Calculus. What if we have a function f(t) and we define a new function F(x) to be the area underneath f(t) between 0 and x. That is,

F(x) = I(0,x) f(t) dt

so the top limit is variable. THE KEY FACT IS THAT

F'(x) = f(x)

Why is this true? We know that

F'(x) = limit as h->0 of [F(x+h) - F(x)]/h

But F(x+h)-F(x) is the area under f(t) between x and x+h. Suppose that between these points f(t) takes minimum value m and maximum value M. Because the width of th interval between x and x+h is h, can you see why (draw a picture at this point)

mh <= F(x+h)-F(x) <= Mh

So m <= [F(x+h) - F(x)]/h <= M. Now as h gets smaller and smaller, x+h gets closer and closer to x, so the maximum and minimum values m and M must get closer and closer to f(x). So [F(x+h) - F(x)]/h is sandwiched between two values both of which get closer and closer to f(x). Therefore it must get closer and closer to f(x). That is,

F'(x) = limit as h->0 of [F(x+h) - F(x)]/h = f(x)

As required. Now we can prove what we wanted to prove. We want to show that

The area under the curve f(x) between a and b is equal to G(b)-G(a), where G(x) differentiates to give f(x), that is G'(x) = f(x).

First note that the area we want is

I(a,b) f(t) dt = I(0,b) f(t) dt - I(0,a) f(t) dt = F(b)-F(a)

Now suppose G(x) is such that G'(x) = f(x). We showed earlier that F'(x) = f(x) so we have F'(x) = G'(x). Therefore F(x) and G(x) differ only by a constant. That is

G(x) = F(x) + C

So finally

G(b) - G(a) = (F(b)+C)-(F(a)+C) = F(b) - F(a)

is the area we want.

I'm able to answer questions if you want.
0
13 years ago
#11
It just does.
so true! Well for Alevel at least.
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