Applied Calculus I Lecture 36
Computing the volume Consider a continuous function over an interval [a, b]. y a b x
Computing the volume Consider a continuous function over an interval [a, b]. y y a b x a b x Imagine we revolve the region bounded by the graph of f and the lines y =, x = a, x = b around the x-axis, creating a solid body called a solid of revolution.
Computing the volume Consider a continuous function over an interval [a, b]. y y a b x a b x Imagine we revolve the region bounded by the graph of f and the lines y =, x = a, x = b around the x-axis, creating a solid body called a solid of revolution. How can we compute the volume of this solid?
Computing the volume Consider a continuous function over an interval [a, b]. y y a b x a b x Imagine we revolve the region bounded by the graph of f and the lines y =, x = a, x = b around the x-axis, creating a solid body called a solid of revolution. How can we compute the volume of this solid? It turns out that the reasoning we used to compute the area under the curve can be extended to figure out how to compute the volume.
y Computing the volume We split the interval [a, b] into many little subintervals [x i, x i+ ] of the same length: x i+ x i = x. a x i x i+ b x
Computing the volume y a x i x i+ b x We split the interval [a, b] into many little subintervals [x i, x i+ ] of the same length: x i+ x i = x. Then the volume of a slab between x i and x i+ is approximately π[f(x i )] 2 x (which is the volume of a circular cylinder of radius f(x i ) and height x).
y Computing the volume We split the interval [a, b] into many little subintervals [x i, x i+ ] of the same length: x i+ x i = x. a x i x i+ b x Then the volume of a slab between x i and x i+ is approximately π[f(x i )] 2 x (which is the volume of a circular cylinder of radius f(x i ) and height x). n Then the total volume can be approximated as π[f(x i )] 2 x, which is a Riemann sum for the function g(x) = π[f(x)] 2. i=
y Computing the volume We split the interval [a, b] into many little subintervals [x i, x i+ ] of the same length: x i+ x i = x. a x i x i+ b x Then the volume of a slab between x i and x i+ is approximately π[f(x i )] 2 x (which is the volume of a circular cylinder of radius f(x i ) and height x). n Then the total volume can be approximated as π[f(x i )] 2 x, which is a Riemann sum for the function g(x) = π[f(x)] 2. This approximation gets better as n, and in the limit gives the exact value. Hence, we obtain the following result. If f is a continuous function on [a, b] then the volume, V, of the solid obtained by rotating the region bounded by the graph of f and the lines y =, x = a, x = b around the x-axis is V = lim n n π[f(x i )] 2 x = i= b a i= π[f(x)] 2 dx
Compute the volume the a solid obtained by rotating the region bounded by f(x) = 2x + and the lines y =, x =, x = 4 around the x-axis.
Compute the volume the a solid obtained by rotating the region bounded by f(x) = 2x + and the lines y =, x =, x = 4 around the x-axis. We can sketch the graph of the function. But there s no need to draw the volume itself. y 4 x
Compute the volume the a solid obtained by rotating the region bounded by f(x) = 2x + and the lines y =, x =, x = 4 around the x-axis. We can sketch the graph of the function. But there s no need to draw the volume itself. y Using the formula we just learned we obtain: 4 x V = 4 π( 2x + ) 2 dx = π 4 = π [ (4 2 + 4) ( 2 + ) ] = 8π (2x + )dx = π (x 2 + x) 4 =
Compute the volume of the solid obtained by rotating the region bounded by f(x) = sec x and the lines y =, x =, x = π around the x-axis. 4 y 2 π 4 x
Compute the volume of the solid obtained by rotating the region bounded by f(x) = sec x and the lines y =, x =, x = π around the x-axis. 4 Again, we can sketch the region. y 2 π 4 x
Compute the volume of the solid obtained by rotating the region bounded by f(x) = sec x and the lines y =, x =, x = π around the x-axis. 4 Again, we can sketch the region. y 2 The volume is V = π 4 π sec 2 xdx = π tan x π 4 = π π 4 x
Average values Suppose we need to compute the average value of a function f over an interval [a, b].
Average values Suppose we need to compute the average value of a function f over an interval [a, b]. If we had only a finite number of values, f(x ), f(x ),..., f(x n ), then the average would be n n i= f(x i )
Average values Suppose we need to compute the average value of a function f over an interval [a, b]. If we had only a finite number of values, f(x ), f(x ),..., f(x n ), then the average would be n f(x i ) n i= We can take these x i to be evenly spaced within [a, b]. Then x = x i+ x i = b a. So, the above average can be written as n n f(x i ) = n f(x i ) x n b a i= i=
Average values Suppose we need to compute the average value of a function f over an interval [a, b]. If we had only a finite number of values, f(x ), f(x ),..., f(x n ), then the average would be n f(x i ) n i= We can take these x i to be evenly spaced within [a, b]. Then x = x i+ x i = b a. So, the above average can be written as n n f(x i ) = n f(x i ) x n b a i= The latter sum is the Riemann sum. As we let n it converges to the average value of f over [a, b]. Therefore, The average value of a continuous function f over an interval [a, b] is given by n lim f(x i ) x = b f(x)dx n b a b a i= i= a
Compute the average value of f(x) = x ln x on the interval [, e].
Compute the average value of f(x) = x ln x on the interval [, e]. The average value is e e x ln xdx. Let s compute the integral. e x ln xdx =
Compute the average value of f(x) = x ln x on the interval [, e]. e The average value is e e e ( ) x 2 x ln xdx = ln xd 2 x ln xdx. Let s compute the integral. =
Compute the average value of f(x) = x ln x on the interval [, e]. e The average value is e e e ( ) x 2 x ln xdx = ln xd 2 x ln xdx. Let s compute the integral. = ( ) x 2 e 2 ln x 2 e x 2 d(ln x) =
Compute the average value of f(x) = x ln x on the interval [, e]. e The average value is e e e ( ) x 2 x ln xdx = ln xd 2 = e2 2 2 e xdx = x ln xdx. Let s compute the integral. = ( ) x 2 e 2 ln x 2 e x 2 d(ln x) =
Compute the average value of f(x) = x ln x on the interval [, e]. e The average value is x ln xdx. Let s compute the integral. e e e ( ) ( ) x 2 x 2 e x ln xdx = ln xd = 2 2 ln x e x 2 d(ln x) = 2 = e2 2 e xdx = e2 2 2 ( ) x 2 e = 2 2
Compute the average value of f(x) = x ln x on the interval [, e]. e The average value is x ln xdx. Let s compute the integral. e e e ( ) ( ) x 2 x 2 e x ln xdx = ln xd = 2 2 ln x e x 2 d(ln x) = 2 = e2 2 e xdx = e2 2 2 ( ) x 2 e = e2 2 2 2 e2 4 + 4 =
Compute the average value of f(x) = x ln x on the interval [, e]. e The average value is x ln xdx. Let s compute the integral. e e e ( ) ( ) x 2 x 2 e x ln xdx = ln xd = 2 2 ln x e x 2 d(ln x) = 2 = e2 2 e xdx = e2 2 2 ( ) x 2 e = e2 2 2 2 e2 4 + 4 = e2 + 4
Compute the average value of f(x) = x ln x on the interval [, e]. e The average value is x ln xdx. Let s compute the integral. e e e ( ) ( ) x 2 x 2 e x ln xdx = ln xd = 2 2 ln x e x 2 d(ln x) = 2 = e2 2 e xdx = e2 2 2 ( ) x 2 e = e2 2 2 2 e2 4 + 4 = e2 + 4 So, the average value is e e 2 + 4 = e2 + 4(e ).226
Supose that the number of items a new worker on an assembly line produces daily after t days on the job is given by I(t) = 45 ln(t + ) Find the average number of items produced daily by this employee over the first five days.
Supose that the number of items a new worker on an assembly line produces daily after t days on the job is given by I(t) = 45 ln(t + ) Find the average number of items produced daily by this employee over the first five days. This average number is 5 5 45 ln (t + )dt =
Supose that the number of items a new worker on an assembly line produces daily after t days on the job is given by I(t) = 45 ln(t + ) Find the average number of items produced daily by this employee over the first five days. This average number is 5 5 45 ln (t + )dt = 9 5 ln (t + )d(t + ) =
Supose that the number of items a new worker on an assembly line produces daily after t days on the job is given by I(t) = 45 ln(t + ) Find the average number of items produced daily by this employee over the first five days. This average number is 5 45 ln (t + )dt = 9 5 [ ] 5 = 9 (t + ) ln (t + ) 9 5 5 ln (t + )d(t + ) = (t + )d(ln (t + )) =
Supose that the number of items a new worker on an assembly line produces daily after t days on the job is given by I(t) = 45 ln(t + ) Find the average number of items produced daily by this employee over the first five days. This average number is 5 45 ln (t + )dt = 9 5 [ ] 5 = 9 (t + ) ln (t + ) 9 5 5 ln (t + )d(t + ) = (t + )d(ln (t + )) = = 54 ln 6 9 5 dt =
Supose that the number of items a new worker on an assembly line produces daily after t days on the job is given by I(t) = 45 ln(t + ) Find the average number of items produced daily by this employee over the first five days. This average number is 5 45 ln (t + )dt = 9 5 [ ] 5 = 9 (t + ) ln (t + ) 9 = 54 ln 6 9 5 5 5 ln (t + )d(t + ) = (t + )d(ln (t + )) = dt = 54 ln 6 9t 5 = 54 ln 6 45 = 9(6 ln 6 5)