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... Enough with derivatives. Let us move on to integrals!
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Integrals
Recall the notion of integration in one variable... b
a
f (x)dx = lim
n→∞ n
- i=1
f (xi)(xi − xi−1) We call this the Riemann integral of f .
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... Enough with derivatives. Let us move on to integrals! 85 / 158 - - PowerPoint PPT Presentation
... Enough with derivatives. Let us move on to integrals! 85 / 158 - - PowerPoint PPT Presentation
... Enough with derivatives. Let us move on to integrals! 85 / 158 Integrals Recall the notion of integration in one variable... b n f ( x ) d x = lim f ( x i )( x i x i 1 ) n a i =1 We call this the Riemann integral of
SLIDE 2
SLIDE 3
Integrals
Why do we like integrals?
◮ They give us the (signed) area/volume under/over a function.
Applications: calculation of areas, arc-lengths, volumes, surface areas etc.
◮ Integration is ”inverse” operation of differentiation.
Applications: calculation of antiderivatives (indefinite integrals)
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SLIDE 4
Fundamental Theorem of Calculus
Theorem: If there exists differentiable function F such that F ′ = f , for an (integrable) function f , then b
a
f (x)dx = F(b) − F(a). Second form of the FTC: d dx x
a
f (t)dt = f (x). BIG IDEA: Differentiation and integration are ”inverse” operations!
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SLIDE 5
Integrals
Let f , g (integrable) functions. Then
◮
- f + gdx =
- f dx +
- gdx
◮
- h(y)f (x)dx = h(y)
- f (x)dx for h (integrable) function.
◮
- f (x)g′(x)dx = f (x)g(x) −
- f ′(x)g(x)dx
- r
b
a
f (x)g′(x)dx = [f (x)g(x)]b
a −
b
a
f ′(x)g(x)dx Integration by parts! ... maybe the most important formula in applied analysis.
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SLIDE 6
Double Integrals
Let f : R2 → R function. We define the double integral of f by
- Ω
f (x, y)dxdy = lim
m→∞ lim n→∞ m
- i=1
n
- j=1
f (xi, yj)(xi − xi−1)(yj − yj−1)
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SLIDE 7
Double Integrals over Rectangles
Let f : [a, b] × [c, d] ⊂ R2 → R (integrable) function. The double integral
- f f is then
- Ω
f (x, y)dxdy = d
c
b
a
f (x, y)dx
- dy =
b
a
d
c
f (x, y)dy
- dx
BIG IDEA: The double integral of a con- tinuous function over a rectangular do- main is equal to the iterated integrals. Remark: We can use either of the two forms (red or blue) depending on which
- ne is more convenient in each particular
case.
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SLIDE 8
Double Integrals over Rectangles
Example 1: Let f (x, y) = cos x sin y. Compute the double integral of f
- ver the rectangle Ω = [0, π
2 ] × [0, π 2 ].
Solution: We have
- π
2
- π
2
cos x sin ydx
- dy = · · · = 1.
Similarly
- π
2
- π
2
cos x sin ydy
- dx = · · · = 1.
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SLIDE 9
Double Integrals over Rectangles
Example 2: Let f (x, y) = x sin y. Find the double integral of f over the rectangle Ω = [0, 1] × [0, π]. Solution: We have
- Ω
x sin ydxdy = π 1 x sin ydx
- dy = · · · = 1.
Similarly
- Ω
x sin ydxdy = 1 π x sin ydy
- dx = · · · = 1.
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SLIDE 10
Double Integrals over Simple Domains
Problem: Let f : D ⊂ R2 → R. What is the double integral of f over D, when D is of the following form? Solution:
- D
f (x, y)dxdy = b
a
φ2(x)
φ1(x)
f (x, y)dy
- dx
Domains of the form D = {(x, y) ∈ R2 : a ≤ x ≤ b, φ1(x) ≤ y ≤ φ2(x)} are called x-simple domains.
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SLIDE 11
Double Integrals over Simple Domains
Example: Let f (x, y) = xy. Compute the double integral of f over the domain bounded by y = √x, the x-axis, x = 1, and x = 3. Solution: D is x-simple. Indeed, D = {(x, y) : 1 ≤ x ≤ 3, 0 ≤ y ≤ √x}. We have 3
1
√x xydydx = · · · = 13 3 .
0.5 1 1.5 2 2.5 3 3.5 4 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
D
x=1 x=3 y=\/x y=0
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