Integrating to Find Area Under a Curve

    • Thread Starter

    Please could someone explain to me WHY calculating a definite integral finds you the area under a curve?

    Wikipedia does a pretty good job of giving a rough idea why it's the case, check out the 'Geometric intuition' bit of

    The answer to this question is a lot easier to understand graphically, but basically you have to imagine that in \int f(x)dx, the f(x) is the height of a rectangle and the dx is some arbitrarily small width of that rectangle. The integral means, in essence, adding up all the rectangles - which amounts to finding the total area under the curve.

    Edit: this is not really super helpful in terms of answering your question, but is roughly relevant and amused me. Why don't I think of things like this?


    (Original post by Claree)
    Please could someone explain to me WHY calculating a definite integral finds you the area under a curve?
    Or, if you want to go more in depth, check out the Riemann Integral
    tbh, it depends what you define an integral to be.

    (Original post by Claree)
    Please could someone explain to me WHY calculating a definite integral finds you the area under a curve?
    I am not sure if i am right, but in my opinion this is the case,

    You would have probably studied calculating area by using trapezium rule, in that rule we use strips.
    In integration, we use infinite strips.

    The area of one strip will be, Area= y \delta x where delta x is very small, almost zero.
    Now summing the strips,
     \displaystyle \sum^b_a y\delta x

    As  \delta x \rightarrow 0 , the summation becomes integration,
     \displaystyle \int^b_aydx

    Hope i am correct.

    Strictly speaking, it's because the integral is what we define the "area under a curve" to be, and the fact it's consistent with the definition of area for simple shapes is what allows this definition to make sense.

    the area (called the Riemann integral) is the limit of the riemann sum over the certain interval - end points (a,b), divided into intervals of equal length b-a/l, with a point c in each interval. Sum is then SIGMA i=0 to n-1 of f(c_i)(x_i+1-x_i)

Write a reply… Reply
Submit reply


Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: February 29, 2012
Which party will you be voting for in the General Election 2017?
Useful resources

Make your revision easier


Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here


How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.