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| By the logic above, a change in ''x'' (or Δ''x'') is the sum of the infinitesimal changes d''x''. It is also equal to the sum of the infinitesimal products of the derivative and time. This infinite summation is integration; hence, the integration operation allows the recovery of the original function from its derivative. It can be concluded that this operation works in reverse; the result of the integral can be differentiated to recover the original function. | | By the logic above, a change in ''x'' (or Δ''x'') is the sum of the infinitesimal changes d''x''. It is also equal to the sum of the infinitesimal products of the derivative and time. This infinite summation is integration; hence, the integration operation allows the recovery of the original function from its derivative. It can be concluded that this operation works in reverse; the result of the integral can be differentiated to recover the original function. |
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| ==Geometric intuition==
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| [[Image:FTC geometric.svg|500px|thumb|right|The area shaded in red stripes can be computed as ''h'' times ''ƒ''(''x''). Alternatively, if the function ''A''(''x'') were known, it could be estimated as ''A''(''x'' + ''h'') − ''A''(''x''). These two values are approximately equal, particularly for small ''h''.]]
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| For a continuous function {{nowrap|1=''y'' = ƒ(''x'')}} whose graph is plotted as a curve, each value of ''x'' has a corresponding area function ''A''(''x''), representing the area beneath the curve between 0 and ''x''. The function ''A''(''x'') may not be known, but it is given that it represents the area under the curve.
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| The area under the curve between ''x'' and ''x'' + ''h'' could be computed by finding the area between 0 and ''x'' + ''h'', then subtracting the area between 0 and ''x''. In other words, the area of this “sliver” would be {{nowrap|''A''(''x'' + ''h'') − ''A''(''x'')}}.
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| There is another way to ''estimate'' the area of this same sliver. ''h'' is multiplied by ƒ(''x'') to find the area of a rectangle that is approximately the same size as this sliver. It is intuitive that the approximation improves as ''h'' becomes smaller.
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| At this point, it is true ''A''(''x'' + ''h'') − ''A''(''x'') is approximately equal to ƒ(''x'')·''h''. In other words, {{nowrap|ƒ(''x'')·''h'' ≈ ''A''(''x'' + ''h'') − ''A''(''x'')}}, with this approximation becoming an equality as ''h'' approaches 0 in the limit.
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| When both sides of the equation are divided by ''h'':
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| : <math>f(x) \approx \frac{A(x+h)-A(x)}{h}.</math>
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| As ''h'' approaches 0, it can be seen that the right hand side of this equation is simply the derivative ''A''’(''x'') of the area function ''A''(''x''). The left-hand side of the equation simply remains ƒ(''x''), since no ''h'' is present.
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| It can thus be shown, in an informal way, that {{nowrap|1 = ƒ(''x'') = ''A''’(''x'')}}. That is, the derivative of the area function ''A''(''x'') is the original function ƒ(''x''); or, the area function is simply the antiderivative of the original function.
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| Computing the derivative of a function and “finding the area” under its curve are "opposite" operations. This is the crux of the Fundamental Theorem of Calculus. Most of the theorem's proof is devoted to showing that the area function ''A''(''x'') exists in the first place.
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| ==Formal statements== | | ==Formal statements== |