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Probably the most important equipment used to demonstrate mathematical results in Calculus certainly is the Mean Benefits Theorem of which states that if f(x) is outlined and is steady on the period [a, b] and is differentiable on (a, b), there is also a number city in the time period (a, b) [which means a b] such that,

f'(c)=[f(b) — f(a)]/(b-a).

Remainder Theorem : Consider a function f(x)=(x-4)^2 + one particular on an length [3, 6]

Remedy: f(x)=(x-4)^2 plus 1, provided interval [a, b]=[3, 6]

f(a)=f(3)=(3-4)^2 + 1= 1+1 =2

f(b)=f(6)=(6-4)^2 plus 1 = 4+1 =5

Using the Mean Value Theory, let us get the derivative at some point vitamins.

f'(c)= [f(b)-f(a)]/(b-a)

=[5-2]/(6-3)

=3/3

=1

Therefore , the derivative at city is 1 ) Let us today find the coordinates of c by just plugging in c in the derivative of this original formula given and place it add up to the result of the Mean Benefit. That gives you,

f(x) = (x-4)^2 plus1

f(c) = (c-4)^2+1

= c^2-8c+16 +1

=c^2-8c+17

f'(c)=2c-8=1 [f'(c)=1]

we get, c= 9/2 which can be the x value in c. Plug in this importance in the first equation

f(9/2) = [9/2 – 4]^2+1= 1/4 plus one = 5/4

so , the coordinates in c (c, f(c)) is (9/2, 5/4)

Mean Worth Theorem meant for Derivatives areas that if perhaps f(x) is mostly a continuous party on [a, b] and differentiable on (a, b) then we have a number c between an important and udemærket such that,

f'(c)= [f(b)-f(a)]/(b-a)

Mean Value Theorem for Integrals

It suggests that in the event that f(x) may be a continuous action on [a, b], then we have a number c in [a, b] such that,

f(c)= 1/(b-a) [Integral (a to b)f(x) dx]

This is the First Mean Value Theorem pertaining to Integrals

From your theorem we can easily say that the standard value from f with [a, b] is attained on [a, b].

Example: Enable f(x) sama dengan 5x^4+2. Determine c, such that f(c) may be the average significance of farreneheit on the time period [-1, 2]

Answer: Using the Mean Value Theorem for the Integrals,

f(c) = 1/(b-a)[integral(a to b) f(x) dx]

The typical value from f over the interval [-1, 2] is given by,

sama dengan 1/[2-(-1)] essential (-1 to 2) [5x^4+2]dx

= one-third [x^5 +2x](-1 to 2)

= 1/4 [ 2^5+ 2(2) – (-1)^5+2(-1) ]

sama dengan 1/3 [32+4+1+2]

= 39/3 sama dengan 13

Seeing as f(c)= 5c^4+2, we get 5c^4+2 = 13, so city =+/-(11/5)^(1/4)

We get, c= next root of (11/5)

Second Mean Value Theorem for the integrals says that, If f(x) is normally continuous on an interval [a, b] in that case,

d/dx Integral(a to b) f(t) dt = f(x)

Example: find d/dx Vital (5 to x^2) sqrt(1+t^2)dt

Solution: Lodging a finance application the second Mean Value Theorem for Integrals,

let u= x^2 which gives us y= integral (5 to u) sqrt(1+t^2)dt

We realize, dy/dx = dy/du. ihr. dx sama dengan [sqrt(1+u^2)] (2x) = twice[sqrt(1+x^4)]