5.1: Area and Estimating with Finite (Riemann) Sums
Important application of calculus:
find the area A under the curve
y = f (x) for a ≤ x ≤ b.
y
y = f (x)
Approximate the area under the
curve by adding the areas of the
rectangles; the i th rectangle has
height f (xi−1 ) and width ∆x.
A ≈ Ln = A1 + · · · + An
Other possible approximations using
sums of rectangles: Take the height
of the i th rectangle to be the value of
the function at the right endpoint or
midpoint of [xi−1 , xi ].
Ln = f (x0 )∆x + · · · + f (xn−1 )∆x
Rn = f (x1 )∆x + · · · + f (xn )∆x
= f (x0 )∆x + · · · + f (xn−1 )∆x
A
x
a
b
Estimate A by subdividing [a, b] into
n subintervals of length ∆x = b−a
n .
y
y = f (x)
Ln is the left sum: xi−1 is the left
endpoint of i th subinterval [xi−1 , xi ].
Increasing n improves the accuracy
of the approximation.
y
y = f (x)
L5
ci =
midpoint of [xi−1 , xi ].
Rn is the right sum and Mn is the
midpoint sum.
y = f (x)
x0
A2
x1
A3
x2
L5 = (02 + .42 + .82 + 1.22 + 1.62 ).4
y = f (x)
y
M5
A4
x3
= 3.52
A5
x4
x5
x
x
x0 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10
Distance problem
If velocity is constant, then
distance = velocity × time
What if velocity is variable?
Subdivide the time interval and
proceed as above.
Ex d: Velocity of a decelerating
vehicle is given in the table in m/s;
find upper, lower estimates of the
distance covered:
t 0
2
4 6
v 36 32 20 0
Upper and lower estimates given by
L3 = (v (0) + v (2) + v (4))∆t
= (36 + 32 + 20)2 = 176 m
R3 = (v (2) + v (4) + v (6))∆t
= (32 + 20 + 0)2 = 104 m
= 1.92
R5 = (.42 + .82 + 1.22 + 1.62 + 22 ).4
L10
x
x0 x1 x2 x3 x4 x5
A1
Ex b: As above but with n = 5.
We have ∆x = 2−0
5 = .4 and
xi = 0, .4, .8, 1.2, 1.6, 2.
xi−1 +xi
:
2
R5
R2 = f (1)∆x + f (2)∆x = 5
M2 = f (.5)∆x + f (1.5)∆x = 2.5
Mn = f (c1 )∆x + · · · + f (cn )∆x
y
Ex a: Estimate area under the graph
of f (x) = x 2 from a = 0 to b = 2
using n = 2 subintervals. Note
∆x = b−a
n = 1, xi = i, i = 0, 1, 2.
L2 = f (0)∆x + f (1)∆x = 1
In the above example we couldn’t
use a midpoint sum because we had
no data at the midpoints.
Ex d: Suppose we know that
v (t) = 36 − t 2 m/s for 1 ≤ t ≤ 6.
Estimate the distance covered using
n = 10. Then ∆t = (6 − 1)/10 = .5
and ti = 1.0, 1.5, 2.0, . . . , 6.0.
L10 = (v (1.0) + v (1.5) + · · · + v (5.5)).5
Midpoint sum is often “best”. Can
use Maple to compute sums.
Average Value
The average value of f (x) over the
interval [a, b] is given by
Area under graph
µ=
b−a
y
y = f (x)
= 99.375 m
M10 = (v (1.25) + · · · + v (5.75)).5
= 108.437 m
Use the Riemann sum applet to
compute cases with n = 20, n = 50.
The actual distance is 108.333 m.
M5 = (.22 + .62 + 1.02 + 1.42 + 1.82 ).4
= 2.64
Ex f: Estimate the average value of
f (x) = cos x over [−π/2, π/2] by
first using a midpoint sum with
n = 3 to estimate the area.
∆x = π3 , xi = − π2 , − π6 , π6 , π2 and
ci = − π3 , 0, π3 .
y
y = cos x
average
µ
= 116.875 m
R10 = (v (1.5) + v (2.0) + · · · + v (6.0)).5
c 1 c2 c 3 c 4 c 5
x
A
x
x
a
b
Ex e: Find the average value of x
over [1, 3]. The area under the curve:
A = 92 − 12 = 4. So the average value
4
of x over [1, 3] is 3−1
= 2.
y
2
1
3
x
−π
2
−π
3
−π
6
π
6
π
3
π
2
�
�π
M3 = f (−π/3) + f (0) + f (π/3)
3
π
2π
= (.5 + 1 + .5) =
3
3
2π/3
2
µest =
=
π/2 − (−π/2)
3
If there is time estimate the average
value using M10 .