MATH 12002 - CALCULUS I 4.4: Average Value & Geometry Professor - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 4.4: Average Value & Geometry Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 12

Average Value Recall that the average value of a function is found as follows: Average Value Let y = f ( x ) be a continuous function. The average value of f on the interval [ a , b ] is � b 1 f ave = f ( x ) dx. b − a a The average value of a positive function can be described in terms of areas. D.L. White (Kent State University) 2 / 12

Average Value Suppose y = f ( x ) is continuous and f ( x ) � 0 for a � x � b , � b 1 with average value f ave = a f ( x ) dx on [ a , b ]. b − a � b We know that a f ( x ) dx is the area under the graph of f on [ a , b ]. � b 1 Since f ave = a f ( x ) dx , this area is also equal to ( b − a ) f ave . b − a Observe that ( b − a ) f ave is the area of the rectangle of height f ave on [ a , b ]. ✻ f f ave ✛ ✲ ❄ a b D.L. White (Kent State University) 3 / 12

Average Value Suppose y = f ( x ) is continuous and f ( x ) � 0 for a � x � b , � b 1 with average value f ave = a f ( x ) dx on [ a , b ]. b − a � b We know that a f ( x ) dx is the area under the graph of f on [ a , b ]. � b 1 Since f ave = a f ( x ) dx , this area is also equal to ( b − a ) f ave . b − a Observe that ( b − a ) f ave is the area of the rectangle of height f ave on [ a , b ]. ✻ f f ave ✛ ✲ ❄ a b D.L. White (Kent State University) 4 / 12

Average Value Hence the average value f ave of y = f ( x ) on [ a , b ] is the number such that the area under the horizontal line y = f ave between x = a and x = b is equal to the area under the graph of y = f ( x ) between x = a and x = b . ✻ ✻ f f f ave f ave ✛ ✲ ✛ ✲ ❄ ❄ a a b b D.L. White (Kent State University) 5 / 12

Average Value Notice that the rectangle and the region under the graph of f have the black shaded region in common. Since the total areas are equal, it follows that the areas of the red shaded regions must be equal. ✻ ✻ f f f ave f ave ✛ ✲ ✛ ✲ ❄ ❄ a a b b D.L. White (Kent State University) 6 / 12

Average Value If you aren’t convinced by those pictures, you can watch the red shaded region move back and forth: ✻ f f ave ✛ ✲ ❄ a b D.L. White (Kent State University) 7 / 12

Average Value If you aren’t convinced by those pictures, you can watch the red shaded region move back and forth: ✻ f f ave ✛ ✲ ❄ a b D.L. White (Kent State University) 8 / 12

Average Value We can also interpret the average value f ave as the number such that: the area above the horizontal line y = f ave and below the graph of y = f ( x ) between x = a and x = b and the area below the horizontal line y = f ave and above the graph of y = f ( x ) between x = a and x = b are equal. ✻ f f ave ✛ ✲ ❄ a b D.L. White (Kent State University) 9 / 12

Mean Value Theorem For Integrals By the Extreme Value Theorem, if y = f ( x ) is continuous on the closed interval [ a , b ], then f has an absolute minimum m and an absolute maximum M on [ a , b ]. Thus m � f ( x ) � M on [ a , b ], and so � b m ( b − a ) � f ( x ) dx � M ( b − a ), a by a comparison property of definite integrals. Dividing by b − a , we get � b 1 f ( x ) dx � M ; m � b − a a that is, m � f ave � M . By the Intermediate Value Theorem, f takes on every y value between m and M , and therefore there is a c with a < c < b such that � b 1 f ( c ) = f ( x ) dx = f ave . b − a a D.L. White (Kent State University) 10 / 12

Mean Value Theorem For Integrals This is the Mean Value Theorem for Integrals: Theorem If y = f ( x ) is a continuous function on [ a , b ] , then there is a number c with a < c < b such that � b 1 f ( c ) = f ( x ) dx. b − a a The theorem says that a continuous function on a closed interval must take on its average (i.e., mean) value on the interval. D.L. White (Kent State University) 11 / 12

Mean Value Theorem For Integrals If y = f ( x ) is continuous on [ a , b ] and F is an antiderivative of f , � b then a f ( x ) dx = F ( b ) − F ( a ), by the Fundamental Theorem of Calculus, and so � b 1 f ( x ) dx = F ( b ) − F ( a ) b − a b − a a and f ( c ) = F ′ ( c ). Hence the Mean Value Theorem for Integrals implies F ( b ) − F ( a ) = F ′ ( c ) b − a for some c with a < c < b . This is the Mean Value Theorem (for F ) that we studied previously. D.L. White (Kent State University) 12 / 12