The Mean Value Theorem for double integrals says that if f is a continuous function on a plane region D that is of type I or II, then there exists a point ( x 0 , y 0 ) in D such that ∬ D f ( x , y ) d A = f ( x 0 , y 0 ) A ( D ) Use the Extreme Value Theorem (14.7.8) and Properly 15.2.11 of integrals to prove this theorem. (Use the proof of the single-variable version in Section 6.5 as a guide.)
The Mean Value Theorem for double integrals says that if f is a continuous function on a plane region D that is of type I or II, then there exists a point ( x 0 , y 0 ) in D such that ∬ D f ( x , y ) d A = f ( x 0 , y 0 ) A ( D ) Use the Extreme Value Theorem (14.7.8) and Properly 15.2.11 of integrals to prove this theorem. (Use the proof of the single-variable version in Section 6.5 as a guide.)
Solution Summary: The author explains the Mean value theorem for double integrals.
The Mean Value Theorem for double integrals says that if f is a continuous function on a plane region D that is of type I or II, then there exists a point (x0, y0) in D such that
∬
D
f
(
x
,
y
)
d
A
=
f
(
x
0
,
y
0
)
A
(
D
)
Use the Extreme Value Theorem (14.7.8) and Properly 15.2.11 of integrals to prove this theorem. (Use the proof of the single-variable version in Section 6.5 as a guide.)
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
a
->
f(x) = f(x) = [x] show that whether f is continuous function or not(by using theorem)
Muslim_maths
Use Green's Theorem to evaluate F. dr, where
F = (√+4y, 2x + √√)
and C consists of the arc of the curve y = 4x - x² from (0,0) to (4,0) and the line segment from (4,0) to
(0,0).
Evaluate
F. dr where F(x, y, z) = (2yz cos(xyz), 2xzcos(xyz), 2xy cos(xyz)) and C is the line
π 1
1
segment starting at the point (8,
'
and ending at the point (3,
2
3'6
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Fundamental Theorem of Calculus 1 | Geometric Idea + Chain Rule Example; Author: Dr. Trefor Bazett;https://www.youtube.com/watch?v=hAfpl8jLFOs;License: Standard YouTube License, CC-BY