warm up – no calculator
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Warm Up – NO CALCULATOR. Let f(x) = x 2 – 2x. Determine the average rate of change of f(x) over the interval [-1, 4]. Determine the value of (Check your answer using your calculator). Mean Value Theorem for Integrals Average Value 2 nd Fundamental Theorem of Calculus. a. c. b. - PowerPoint PPT PresentationTRANSCRIPT
Warm Up – NO CALCULATORLet f(x) = x2 – 2x.1) Determine the average rate of change of
f(x) over the interval [-1, 4].
2) Determine the value of(Check your answer using your calculator)
4
1
f (x)dx
Mean Value Theorem for Integrals
Average Value
2nd Fundamental Theorem of Calculus
Mean Value Theorem for Integrals
If f is continuous on [a,b] then there is a certain point (c, f(c)) between a and b so if you draw a rectangle whose length is the interval [a,b] and
whose height is f(c), the area of the rectangle will be exactly the area beneath the function on [a,b].
a bc
In other words…
*If f is continuous on [a,b], then there exists a number c in the open interval (a,b) such that .
( ) ( )( )b
af x dx f c b a
Area under the curve from a to b
Area of the rectangle formed
Example 1: Find the value of f(c) guaranteed by
MVT for integration for the function f(x) = x3 – 4x2 + 3x + 4 on [1,4]
Explain the relationship of this value to the graph of f(x)?
Example 2Find the value of f(c) guaranteed by MVT for integrals on the interval [1,9] for
1f(x) x
2
The f(c) value you found in both examples is
called the average value of f.
Solving for f(c) gives the formula for average value.
1( ) ( )
b
a
Average Value f c f x dxb a
Example 3: Find the average value of f(x) = 3x2 – 2x on the interval [1,4] and all values of x in the interval for which the function equals its average value.
Taking the derivative of a definite integral whose lower bound is a number and whose upper bound contains a variable.
f (x)
a
dg(y)dy
dx
The long way…x
2
d3t dt
dx
4x
2
d3t dt
dx
The 2nd Fundamental Theorem of Calculus:If f(x) is continuous and differentiable,
f (x)
a
dg(y)dy g(f(x))f ' (x)
dx
Here’s what you REALLY do…
x
2
d3t dt
dx
4x
2
d3t dt
dx
Your turn…
2
3
x2
x
0
12
t
d1) (y 6y)dy
dx
d2) (cos )d
dx
d3) 2x 3dx
dt
If 2x
2t 3
1
F x dt , then F' 3
Let f be defined on the closed interval [-5,5]. The graph of f consisting of two line segments and two semicircles, is shown above.
f
1
5
( ) f x dx5
5
( )
f x dx
Find g(2)
1
( ) ( )x
g x f t dt
f
Find g’(2)
Find g”(2)
Let g be the function given by
1
( )x
f t dt g(x)=
f
On what intervals, if any, is g increasing?
Find the x-coordinate of each point of inflection of the graph g on the open interval (-5,5). Justify your answer.
1
( )x
f t dt g(x)=
f
Find the average rate of change of g on the interval [-5,5].