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# 香港六合彩

## on May 27, 2008

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## 香港六合彩Presentation Transcript

• Areas of Domains and Definite Integrals Area under a Parabola Definition of the Area of Certain Domains Area under a Parabola Revisited Integrals and Antiderivatives
• Estimate Areas Consider the problem of determining the area of the domain bounded by the graph of the function x 2 , the x -axis, and the lines x =0 and x =1. As the number n of the approximating rectangles grows, the approximation gets better. We determine the area by approximating the domain with thin rectangles for which the area can be directly computed. Letting these rectangles get thinner, the approximation gets better and, at the limit, we get the area of the domain in question. 1 0 y = x 2
• Estimate Areas (2) Let A denote the actual area of the domain in question. Clearly s n < A for all n . Lower est. s n Upper est. S n Height of the k th rectangle. Length of the bottom.
• Estimate Areas (3) This can be computed directly using a previously derived formula for the sum of squares. Solution follows.
• Estimate Areas (4) The blue area under the curve y = x 2 over the interval [0,1] equals 1/3. 1 Conclude 0 y = x 2
• Definition of the Area of a Domain under the Graph of a Function Theorem The previous considerations were based on some intuitive idea about areas of domain. We make that precise in the statement of the following result. Definition The common value of these limits is the area of the domain under the graph of the function f and over the interval [ a , b ].
• The Integral (1) Theorem Definition Notation The common value of these limits is the integral of the function f over the interval [ a , b ].
• The Integral (2) Remark This means that the points x k may be freely selected from the intervals [ a + ( k -1)/ n , a + k / n ]. The limit of the sum does not depend on the choice of the points x k .
• The Integral (3) Remark
• Examples (1) Example 1 Solution By the Definition Conclude Now use the formula for the sum of squares
• Examples (2) Example 2 Solution Conclude The red area under the graph of x 2 over [ a , b ] equals the area over [0, b ] minus the area over [0, a ]. a 0 b
• Integrals and Antiderivatives Rewriting the previous result in the following form we observe that the integral defines the function Direct differentiation yields F’( x ) = x 2 , i.e., the function F is an antiderivative of the function f( x ) = x 2 . Theorem A proof of this result will be presented later.