# If f(x)  0 for a  x  b, then b f(x) dx a is defined to be the area of the region under the graph of f(x) on the interval [a,b]. The precise definition.

## Presentation on theme: "If f(x)  0 for a  x  b, then b f(x) dx a is defined to be the area of the region under the graph of f(x) on the interval [a,b]. The precise definition."— Presentation transcript:

If f(x)  0 for a  x  b, then b f(x) dx a is defined to be the area of the region under the graph of f(x) on the interval [a,b]. The precise definition of this integral as the area is typically done by defining the integral as a limit of sums; this leads to the Fundamental Theorem of Calculus, which states that such integrals can be calculated using anti-differentiation. The definition of this integral can then be extended so that f(x) may possibly take on negative values (and areas below the x axis will be negative).

If f(x,y)  0 for all (x,y) in a region D of the xy plane, then the double integral is defined to be the volume under the graph of f(x,y) on the region D. When the region D is a rectangle R = [a,b]  [c,d], then the volume is that of the following solid: (b,c) z (a,c) (a,d) (b,d) z = f(x,y) y x f(x,y) dx dy= f(x,y) dA DD

ExampleSuppose f(x,y) = k where k is a positive constant, and let R be the rectangle [a,b]  [c,d]. Find (b,c) z (a,c) (a,d) (b,d) f(x,y) = k Since the integral is the volume of a rectangular box with dimensions b – a, d – c, and k, then the integral must be equal to y x k(b – a)(d – c). f(x,y) dA. R ExampleLet R be the rectangle [0,1]  [0,1]. Find (1,1,0) x y z Since the integral is the volume of half of a unit cube, then the integral must be equal to f(x,y) = (1 – x) (1,0,0) (0,0,1)(0,1,1) (0,1,0) 1/2. (1 – x) dA. R

Cavalieri’s Principle : Suppose a solid is sliced by a series of planes parallel to the yz plane, each labeled P x, and the solid lies completely between P a and P b. z y x A(x)A(x) PxPx Let A(x) be the area of the slice cut by P x. If  x represents the distance between any pair of parallel planes, then A(x)  x is approximately the volume of Summing the approximations A(x)  x is an approximation to the portion of the solid between P x and its successor. the volume of the solid. Taking the limit of the sum as  x  0, we find that the solid’s volume is b A(x) dx a

Apply Cavalieri’s principle to find the volume of the following cone: length = 12 length = 13 circle of radius z x y 2.5 y = 5 — x 12 For 0  x  12, the slice of the cone at x parallel to the yz plane is a circle with diameter The area of the slice of the cone at x is A(x) = 5 — x. 12 25 —–  x 2. 576 The volume of the cone is A(x) dx = a b 0 12 25 —–  x 2 dx = 576 25 

To apply Cavalieri’s principle to the double integral of f(x,y) over the rectangle R = [a,b]  [c,d], we first find, for each value of x, the area A(x) of a slice of the solid formed by a plane parallel to the yz plane: (b,c) x y z (a,c) (a,d) (b,d) z = f(x,y) A(x) = Consequently, d f(x,y) dy c f(x,y) dA = R b A(x) dx = a b d f(x,y) dydx. ac By reversing the roles of x and y, we may write f(x,y) dA = R d b f(x,y) dxdy. ca

These equations turn out to be valid even when f takes on negative values. The integrals on the right side of each equation are called iterated integrals (and sometimes the brackets are deleted). ExampleLet R be the rectangle [–1,1]  [0,1]. Find (x 2 + 4y) dA. R (x 2 + 4y) dA = R (x 2 + 4y) dy dx = 0 11 –1–1 x 2 y + 2y 2 dx = y = 0 1 1 –1–1 (x 2 + 2) dx = – 1 1 x 3 /3 + 2x = x = –1 1 14 — 3 Alternatively, one could find that (x 2 + 4y) dx dy = 0 11 –1–1 14 — 3

ExampleLet R be the rectangle [0,  /2]  [0,  /2]. Find (sin x)(cos y) dA. R (sin x)(cos y) dA = R (sin x)(cos y) dy dx = 0  /2 0 (sin x)(sin y) dx =  /2 0 y = 0  /2 (sin x) dx = 0  /2 – cos x = x = 0  /2 1 Alternatively, one could find that (sin x)(cos y) dx dy = 0  /2 0 1

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