MEAN VALUE THEOREM FOR INTEGRALS EXAMPLES

If f is continuous on [a, b], then at some point c in [a, b],

mean-value-theorem-of-integral

The graphical rectangular interpretation of the Mean value theorem for Definite Integrals is that:

If f is continuous on [a, b], then at some point c in [a, b] there is a rectangle with height f(c), and length b – a, such as the area of the rectangle equals the area under the curve f(x) on the interval [a, b]

For each problem, find the values of c that satisfy the Mean Value Theorem for Integrals.

Problem 1 :

f(x) = -x22+x+32 ; [-3,1]
mean-value-theorem-of-integral-q1

Solution :

f(c) = f(c) = = = = = = = = =

Equating the value derived from mean value theorem for integrals to f(x), we get

-c22+c+32 =-23-c22+c+32+23=0-c22+c+136=0-3c2+6c+136=0-3c2+6c+13 = 0x = -b±b2-4ac2aa = -3, b = 6, c = 13c = -6±62-4(-3)(13)2(-3)c = -6±36+156-6c = -6±192-6c = -6±83-6c = -3±43-3c = -3+43-3,c= -3-43-3(not in the given interval)c = 3-433

Problem 2 :

f(x) = 4/x2 ; [-4, -2]

Solution :

Here a = -4, b = -2

Equating the value derived from mean value theorem for integrals to f(x), we get

4c2=12

For each problem, find the average value of the function over the given interval. Then, find the values of c that satisfy the Mean Value Theorem for Integrals.

Problem 3 :

f(x) = −x + 2; [−2, 2]

Solution :

Finding average value of the function :

Finding the value of c that lies in the given interval :

So, the value of c is 0, which lies in the interval.

Problem 4 :

Solution :

Finding average value of the function :

Finding the value of c, using mean value theorem :

So, the value of c is 3.8, which lies in the interval.

Problem 5 :

Solution :

Finding average value of the function :

Finding the value of c, using mean value theorem :

f(x)=4(2x+6)2f(c)=4(2c+6)24(2c+6)2=16(2c+6)2=24(2c+6)=±242c+6=26 and 2c+6=-262c =26-6 and 2c=-26-6c =6-3 (doesn't lie in the interval)and c=-6-3

Problem 6 :

Solution :

Finding average value of the function :

Finding the value of c, using mean value theorem :

f(x)=-x2-8x-17f(c)=-c2-8c-17-x2-8x-17 =-2-x2-8x-17+2 =0-x2-8x-15 =0x2+8x+15 =0(x+3)(x+5)= 0x = -3 and x = -5

Recent Articles

  1. Finding Range of Values Inequality Problems

    May 21, 24 08:51 PM

    Finding Range of Values Inequality Problems

    Read More

  2. Solving Two Step Inequality Word Problems

    May 21, 24 08:51 AM

    Solving Two Step Inequality Word Problems

    Read More

  3. Exponential Function Context and Data Modeling

    May 20, 24 10:45 PM

    Exponential Function Context and Data Modeling

    Read More