fundamental theorem of calculus for Riemann integration

In this entry we discuss the fundamental theorems of calculus for Riemann integration.

- Let $f$ be a Riemann integrable function on an interval $[a,b]$ and $F$ defined in $[a,b]$ by $F(x)=\\int_{a}^{x}f(t)\\,dt+k$ , where $k\\in\\mathbb{R}$ is a constant. Then $F$ is continuous in $[a,b]$ and $F^{\\prime}=f$ almost everywhere (http://planetmath.org/MeasureZeroInMathbbRn).

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- Let $F$ be a continuous function in an interval $[a,b]$ and $f$ a Riemann integrable function such that $F^{\\prime}(x)=f(x)$ except at most in a finite number of points $x$ . Then $F(x)-F(a)=\\int_{a}^{x}f(t)\\,dt$ .

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0.1 Observations

Notice that the second fundamental theorem is not a converse of the first. In the first we conclude that $F^{\\prime}=f$ except in a set of measure zero (http://planetmath.org/MeasureZeroInMathbbRn), while in the second we assume that $F^{\\prime}=f$ except in a finite number of points. In fact, the two theorems can never be the converse of each other as the following example shows:

Example : Let $F$ be the devil staircase function, defined on $[0,1]$ . We have that

•
$F$ is continuous in $[0,1]$ ,
•
$F^{\\prime}=0$ except in a set of (this set must be contained in the Cantor set),
•
$f:=0$ is clearly a Riemann integrable function and $\\int_{0}^{x}0\\,dt=0$ .

Thus, $F(x)\eq\\int_{0}^{x}F^{\\prime}(t)\\,dt$ .

This leads to the question: what kind functions $F$ can be expressed as $F(x)=F(a)+\\int_{a}^{x}g(t)\\,dt$ , for some function $g$ ? The answer to this question lies in the concept of absolute continuity (http://planetmath.org/AbsolutelyContinuousFunction2) (a which the devil staircase does not possess), but for that a more general of integration must be developed (the Lebesgue integration (http://planetmath.org/Integral2)).

单词	FundamentalTheoremOfCalculusForRiemannIntegration
释义	fundamental theorem of calculus for Riemann integration In this entry we discuss the fundamental theorems of calculus for Riemann integration. - Let $f$ be a Riemann integrable function on an interval $[a,b]$ and $F$ defined in $[a,b]$ by $F(x)=\\int_{a}^{x}f(t)\\,dt+k$ , where $k\\in\\mathbb{R}$ is a constant. Then $F$ is continuous in $[a,b]$ and $F^{\\prime}=f$ almost everywhere (http://planetmath.org/MeasureZeroInMathbbRn). $\\,$ - Let $F$ be a continuous function in an interval $[a,b]$ and $f$ a Riemann integrable function such that $F^{\\prime}(x)=f(x)$ except at most in a finite number of points $x$ . Then $F(x)-F(a)=\\int_{a}^{x}f(t)\\,dt$ . $\\,$ 0.1 Observations Notice that the second fundamental theorem is not a converse of the first. In the first we conclude that $F^{\\prime}=f$ except in a set of measure zero (http://planetmath.org/MeasureZeroInMathbbRn), while in the second we assume that $F^{\\prime}=f$ except in a finite number of points. In fact, the two theorems can never be the converse of each other as the following example shows: Example : Let $F$ be the devil staircase function, defined on $[0,1]$ . We have that • $F$ is continuous in $[0,1]$ , • $F^{\\prime}=0$ except in a set of (this set must be contained in the Cantor set), • $f:=0$ is clearly a Riemann integrable function and $\\int_{0}^{x}0\\,dt=0$ . Thus, $F(x)\eq\\int_{0}^{x}F^{\\prime}(t)\\,dt$ . This leads to the question: what kind functions $F$ can be expressed as $F(x)=F(a)+\\int_{a}^{x}g(t)\\,dt$ , for some function $g$ ? The answer to this question lies in the concept of absolute continuity (http://planetmath.org/AbsolutelyContinuousFunction2) (a which the devil staircase does not possess), but for that a more general of integration must be developed (the Lebesgue integration (http://planetmath.org/Integral2)).
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