MATH 1952/51 Class 11 Minh-Tam Trinh University of Chicago 217-1-23
1 Discuss the midterm. 2 Review Riemann sums for functions of one variable. 3 Riemann sums for multivariable functions. 4 Double integrals and their properties. 5 Fubini s theorem.
Riemann Sums in One Variable Why do you need to know integration?
Riemann Sums in One Variable Why do you need to know integration?. A Mathematical Model for the Determination of Total Area Under Glucose Tolerance and Other Metabolic Curves. Diabetes Care, 17(2), Feb. 1994, 152-154.
Riemann Sums in One Variable Why do you need to know integration?. A Mathematical Model for the Determination of Total Area Under Glucose Tolerance and Other Metabolic Curves. Diabetes Care, 17(2), Feb. 1994, 152-154.
The author rediscovered the Trapezoid Rule. (I do not fault the author as much as the peer reviewers who let the article slip by, and the 36 citations it receives on Google.)
Left / right / overestimating / underestimating Riemann sums. Which is which?
To find b f (x) dx, we partition the interval [a, b] into subintervals. a
To find b f (x) dx, we partition the interval [a, b] into subintervals. a In the kth interval, we choose a point x k. If k is the length of the kth interval, then (1) b a f (x) dx f (x k ) k. k
Anatomy of k f (x k ) k :
Anatomy of k f (x k ) k : 1 k is the width of the kth bar.
Anatomy of k f (x k ) k : 1 k is the width of the kth bar. 2 f (x k ) is the height of the kth bar.
Anatomy of k f (x k ) k : 1 k is the width of the kth bar. 2 f (x k ) is the height of the kth bar. 3 k is a shorthand that means sum up f (x k ) k over all intervals.
Example Let f (x) = x 2 4. Approximate 2 f (x) dx, using a Riemann sum over intervals of length.5 and the midpoint of each interval.
Example Let f (x) = x 2 4. Approximate 2 f (x) dx, using a Riemann sum over intervals of length.5 and the midpoint of each interval. The partition is <.5 < 1 < 1.5 < 2.
Example Let f (x) = x 2 4. Approximate 2 f (x) dx, using a Riemann sum over intervals of length.5 and the midpoint of each interval. The partition is <.5 < 1 < 1.5 < 2. The midpoints are.25,.75, 1.25, 1.75.
Example Let f (x) = x 2 4. Approximate 2 f (x) dx, using a Riemann sum over intervals of length.5 and the midpoint of each interval. The partition is <.5 < 1 < 1.5 < 2. The midpoints are.25,.75, 1.25, 1.75. The Riemann sum is (2) f (.25)(.5) + f (.75)(.5) + f (1.25)(.5) + f (1.75)(.5) = 5.28125.
Example Let f (x) = x 2 4. Approximate 2 f (x) dx, using a Riemann sum over intervals of length.5 and the midpoint of each interval. The partition is <.5 < 1 < 1.5 < 2. The midpoints are.25,.75, 1.25, 1.75. The Riemann sum is (2) f (.25)(.5) + f (.75)(.5) + f (1.25)(.5) + f (1.75)(.5) = 5.28125. The actual integral is (3) 2 (x 2 4) dx = ( 1 3 x3 4x) 2 5.33.
Example Let f (x) = x 2 4. Approximate 2 f (x) dx, using a Riemann sum over intervals of length.5 and the midpoint of each interval. The partition is <.5 < 1 < 1.5 < 2. The midpoints are.25,.75, 1.25, 1.75. The Riemann sum is (2) f (.25)(.5) + f (.75)(.5) + f (1.25)(.5) + f (1.75)(.5) = 5.28125. The actual integral is (3) 2 (x 2 4) dx = ( 1 3 x3 4x) Draw this. What does the negative sign mean? 2 5.33.
Riemann Sums in Two Variables
b f (x) dx a signed area between y = f (x) and x-axis d c b a g(x, y) dx dy signed volume between z = g(x, y) and (x, y)-plane subintervals of [a, b] subrectangles of [a, b] [c, d] of length k of area j,k area of bar volume of prism { }} {{ }} { k f (x k ) k j,k f (x j, y k ) j,k
Example Estimate the integral of f (x, y) = cos x + cos y over the region [, 2π] [, 2π], using squares of side length π/2.
Example Estimate the integral of f (x, y) = cos x + cos y over the region [, 2π] [, 2π], using squares of side length π/2. Sixteen squares: Draw them!
Example Estimate the integral of f (x, y) = cos x + cos y over the region [, 2π] [, 2π], using squares of side length π/2. Sixteen squares: Draw them! One possible Riemann sum: (4) f (, ) π2 4 + f (, π 2 ) π2 π2 4 + f (, π) 4 + f (, 3π 2 ) π2 4 + f ( π π2 2, ) 4 + f ( π 2, π 2 ) π2 4 + f ( π π2 2, π) 4 + f ( π 2, 3π 2 ) π2 4 + f (π, ) π2 4 + f (π, π 2 ) π2 π2 4 + f (π, π) 4 + f (π, 3π 2 ) π2 4 + f ( 3π 2 π2, ) 4 + f ( 3π 2, π 2 ) π2 4 + f ( 3π π2 2, π) 4 + f ( 3π 2, 3π 2 ) π2 4 = (2 + 1 + + 1 + 1 + 1 + + 1 2 1 + 1 + 1 + ) π2 4 =. This Riemann sum used the southwest corners. What would the Riemann sum be if we used the midpoints?
Double Integrals When we find the partial derivative x f (x, y), we treat y as a constant.
Double Integrals When we find the partial derivative x f (x, y), we treat y as a constant. When we find the integral b f (x, y) dx, we again treat y as a a constant.
Double Integrals When we find the partial derivative x f (x, y), we treat y as a constant. When we find the integral b f (x, y) dx, we again treat y as a a constant. Example Integrate with respect to x: 1 (x + y) dx. 1 1 y dx. π y sin x dx.
Double Integrals When we find the partial derivative x f (x, y), we treat y as a constant. When we find the integral b f (x, y) dx, we again treat y as a a constant. Example Integrate with respect to x: 1 (x + y) dx. ( 1 2 x2 + xy) 1 x= = 1 2 + y 1 1 y dx. π y sin x dx.
Double Integrals When we find the partial derivative x f (x, y), we treat y as a constant. When we find the integral b f (x, y) dx, we again treat y as a a constant. Example Integrate with respect to x: 1 (x + y) dx. ( 1 2 x2 + xy) 1 x= = 1 2 + y 1 1 y dx. (xy) 1 x= 1 = 2y π y sin x dx.
Double Integrals When we find the partial derivative x f (x, y), we treat y as a constant. When we find the integral b f (x, y) dx, we again treat y as a a constant. Example Integrate with respect to x: 1 (x + y) dx. ( 1 2 x2 + xy) 1 x= = 1 2 + y 1 1 y dx. (xy) 1 x= 1 = 2y π y sin x dx. ( y cos x) π x= = 2y
Example Compute 3 3 5 1 xy2 dx dy.
Example Compute 3 3 5 1 xy2 dx dy. Since y 2 is a constant in the inner integral, we can take it outside: (5) 3 5 3 1 xy 2 dx dy = 3 3 y 2 ( 5 1 ) x dx dy.
Example Compute 3 3 5 1 xy2 dx dy. Since y 2 is a constant in the inner integral, we can take it outside: (5) 3 5 3 1 xy 2 dx dy = 3 3 y 2 ( 5 1 ) x dx Then 5 x dx is a constant in the outer integral! So we can take it 1 outside too: 3 ( 5 ) ( 5 ) ( 3 ) (6) y 2 x dx dy = x dx y 2 dy. 3 1 1 3 dy.
Example Compute 3 3 5 1 xy2 dx dy. Since y 2 is a constant in the inner integral, we can take it outside: (5) 3 5 3 1 xy 2 dx dy = 3 3 y 2 ( 5 1 ) x dx Then 5 x dx is a constant in the outer integral! So we can take it 1 outside too: 3 ( 5 ) ( 5 ) ( 3 ) (6) y 2 x dx dy = x dx y 2 dy. 3 1 1 3 Finally, 5 1 x dx = 52 1 2 2 = 12 and 3 3 y2 dy = 33 ( 3) 3 3 = 18, so the answer is (12)(18) = 156. dy.
Example Find 2π 2π (cos x + cos y) dx dy.
Example Find 2π 2π (cos x + cos y) dx dy. Integrate one variable at a time. First, (7) 2π 2π (cos x + cos y) dx dy = = 2π 2π (sin x + x cos y) 2π x= dy (2π cos y) dy.
Example Find 2π 2π (cos x + cos y) dx dy. Integrate one variable at a time. First, (7) Next, 2π 2π (cos x + cos y) dx dy = = 2π 2π (sin x + x cos y) 2π x= dy (2π cos y) dy. (8) 2π (2π cos y) dy = 2π (sin y) 2π y= =. Notice how the notation 2π, etc., helps me keep track. x=
Fubini s Theorem Theorem (Fubini) If f is continuous or almost continuous, then (9) d b c a f (x, y) dx dy = b d a c f (x, y) dy dx for any rectangle [a, b] [c, d] in the (x, y)-plane.
Fubini s Theorem Theorem (Fubini) If f is continuous or almost continuous, then (9) d b c a f (x, y) dx dy = b d a c f (x, y) dy dx for any rectangle [a, b] [c, d] in the (x, y)-plane. Fubini : integration :: Clairaut : differentiation.
Example (Stewart, 15.1, Example 6) Find 2 1 π y sin(xy) dy dx.
Example (Stewart, 15.1, Example 6) Find 2 1 π y sin(xy) dy dx. Integrating y sin(xy) w.r.t. y is hard. But integrating it w.r.t. x is easy.
Example (Stewart, 15.1, Example 6) Find 2 1 π y sin(xy) dy dx. Integrating y sin(xy) w.r.t. y is hard. But integrating it w.r.t. x is easy. Use Fubini to switch the order: (1) π 2 1 y sin(xy) dx dy = = π π ( cos(xy)) 2 x=1 dy ( cos(2y) + cos(y)) dy = ( 1 2 sin(2y) + sin(y)) π =.