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    Please could someone explain to me WHY calculating a definite integral finds you the area under a curve?
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    Wikipedia does a pretty good job of giving a rough idea why it's the case, check out the 'Geometric intuition' bit of http://en.wikipedia.org/wiki/Fundame...em_of_calculus
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    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?


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    (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.
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    (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.
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    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.
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    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)

    basically:f(c_0)(x_1-x_0)+f(c_1)(x_2-x_1)+f(c_2)(x_3-x_2)+.....+f(c_n-1)(x_n-x_n-1)
 
 
 
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