(1) Consider the function y = f (x) = ex on the interval [0, 1].
(a) Find the area under the graph of this function over this interval using the Fundamental
Theorem of Calculus.
(b) Subdivide the interval [0, 1] into 4 equal subintervals of width 0.25 and find an approximation to the area of this region using four rectangular strips. Use the left hand
rule.
Solution: (a) Area =
Z
1
ex dx = ex |10 = e − 1.
0
(b)Using the left hand end points of the subintervals (”lower sums”) we get the aprroximation
Area ≈
1 0
e + e0.25 + e0.5 + e0.75 ≈ 1.51.
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If use the right hand end points (”upper sums”) we would get 1.942. hw4d.eps
(2) Find the area between the graphs of y = f (x) = 2 − 3x2 and y = −x2 .
Solution: The points of intersection of these two curves satisfy 2 − 3x2 = −x2 , so that
2 = 2x2 , x2 = 1, x = ±1. The region is symmetric about the y axis so the area is
1
Z 1
Z 1
3
8
x
1
2
2
2
= .
2
((2 − 3x ) − (−x ))dx = 2
(2 − 2x )dx = 4 x −
=4 1−
3
3
3
0
0
0
(3)
(a) Find the area enclosed between the graphs of the functions y = f (x) = xn and the
straight line y = x in the first quadrant.
(b) Use your answer in part (a) to find the area between the graphs of the functions
y = f (x) = xn , y = g(x) = x1/n in the first quadrant. (Hint: what is the relationship
between these two functions and what sort of symmetry do their graphs satisfy?)
Solution: (a) For n = 1, 2 · · · the graphs y = xn and y = x intersect at x = 0, 1. Thus the
area is
Z
1
(x − x )dx =
n
0
x2
xn+1
−
2
n+1
1
= 1 − 1 = n−1 .
2 n+1
2(n + 1)
0
(Note that we are considering positive values of n and that for n = 1 the area is zero.
(b) The graphs y = xn and y = x1/n intersect at x = 0, 1. They form an inverse function
pair and therefore are symmetric about the line y = x. Thus the area between them is
twice the answer in (a).
1
(4) Find the exact area of the piece of land which is bounded by the y axis on the west,
the x axis in the south, the lake described by the function y = f (x) = 100 + (x/100)2 in
the north and the line x = 1000 in the east. (Distances are given in feet.)
Solution: The area is
Z
1000
0
1000
1000
x 2 x3 4
+
= 105 .
100 +
dx = 100x
100
30000 0
3
0
(5) Find the area between the graphs of the functions y = f (x) = 2x, y = g(x) = 1 + x2
and the y axis.
Solution: The graphs intersect tangentially at x = 1, and so the area is
Z
Z
1
(1 + x − 2x)dx =
2
0
1
1
(x − 1)3 1
(x − 1) dx =
=
.
3
3
0
2
0
(6) Find the area of the finite plane region bounded by the parabolas y = 6 − x2 and
y = x2 − 4x.
Solution: Here is the picture:
2
The two curves intersect at x = −1 and x = 3. The area is
Z 3
Z 3
2
2
(6 − x ) − (x − 4x) dx =
6 + 4x − 2x2 dx
−1
−1
3
= 6x + 2x2 − 2x3 /3 −1
= 64/3
(7) (a) Find the area enclosed between the graphs of the functions y = x3 and the function
y = x1/3 in the first quadrant. (b) Find the area enclosed between the graphs of the
functions y = x3 and y = x in the first quadrant.
Solution: Here is the picture:
The anaswer to (b) is
Z
0
1
1
x − x3 dx = x2 /2 − x4 /4 0 = 1/4
By a symmetry argument about functional inverses, the answer to (a) is twice that.
√
(8) (a) Estimate the area under the graph of f (x) = x2 + 1 from x = −1 to x = 2 using
three rectangles and right endpoints. Then improve your estimate by using 6 rectangles.
Sketch the graph and the rectangles.
(b) Repeat part (a) using left endpoints.
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(c) Repeat part (a) using midpoints.
(d) From your sketches in parts (a), (b) and (c), which appears to be the best estimate
you can make without doing any more nasty arithmetic?
Solution: Here are the pictures:
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The midpoint rule clearly works better than either of the other two, and of course the finer
the slices the better it works.
(9) Determine a function f (x) and an interval on the x-axis such that the expression shown
below is equal to the area under the graph of f (x) over the given interval:
lim
n→∞
n
X
p
(3/n) 1 + (3i/n)
i=0
What do the single summands that appear in this expression represent?
Solution: ¡¡ No solution. ¿¿
5
(10) (a) Compute the area of the region y ≥ x2 , 1 ≤ y ≤ 2. (Hint: reduce to an area you
already know.)
(b) Let V be the volume of the solid obtained when the same region is rotated rigidly
about the y-axis. Compute V by first approximating V by appropriate slices and then
taking the limit.
Solution: Here is the picture:
The area is therefore
√
Z
√
√
(3)(2 3) − (2)(2 2) − 2 √
2
3
3 √3
x dx = 6 3 − 4 2 − 2 x /3 √2
2
√
√
As for (b), we just slice the volume up into horizontal disks. The volume is
Z
Z
3
3
2
πx dy =
2
√
π( y)2 dy
2
Z
=
3
πy dy
2
3
= πy 2 /2 2 = π(4.5 − 2) = 2.5π
(11) For each of the following functions, find the area under the graph of the function on
the interval [0, 10]. (a) y = 1, (b) y = 1 + x, (c) y = 1 + x2 .
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Solution: (a)
Z
10
1 dx = 10
0
(b)
Z
10
(1 + x) dx = 10 + 100/2 = 60
0
(c)
Z
10
(1 + x2 ) dx = 10 + 1000/3 = 343 1/3
0
(12) Compute the area of the finite region bounded by the curves x = 0, x = 3, y = x + 2,
y = x2 .
Solution: The two curves intersect at −1, 2. The total area is
Z
Z
3
(x + 2) − x dx +
2
−1
2
2
3
x2 − (x + 2) dx = x2 /2 + 2x − x3 /3 −1 + x3 /3 − x2 /2 − 2x 2
= 2 + 4 − 8/3 − 1/2 + 2 + 1/3 + 27/3 − 9/2 − 6 − 8/3 + 2 + 4
=7
(13) Find the area between the two curves y = 1 − x and y = x2 − 1 for x > 0.
Solution: The graphs look like this:
They intersect at x = −2, x = 1. So the area is
Z
1
0
1
(1 − x) − (x2 − 1) dx = 2x − x2 /2 − x3 /3 0
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(14) The growth rate of two children (in gm/day) is shown in this figure:
0 years
1 year
Suppose both have the same weight at time t = 0. Which one is larger at 6 months? at 1
year?
Solution: The weight of each child is the area under the graph of its growth rate. No
quantitative estimates are given, so this question just involves a visual comparison of areas.
The child whose graph starts out on top is larger for a long time, certainly more than six
months. But by one year the other has caught up and passed it.
(15) Two cars, labeled 1 and 2, start side by side and accelerate from rest.
Car 2The figure
shows a graph of their velocity functions. (The vertical scale is in multiples of 10 km per
hour, the horizontal one in minutes. These cars are pretty slow.)
(a) At what time(s) do the cars have equal velocities?
(b) At what time(s) do the cars have equal accelerations?
(c) Which car is ahead after one minute?
(d) Which car is ahead after 3 minutes?
(e) When does one car overtake the other?
Solution: (a) After one minute.
(b) About t = 0.5 and t = 3.
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Car 1
(c) Car #1.
(d) Car #2.
I(t)
(e) About t = 2.
C(t)
(16) The level of glucose in the blood depends on the rate of intake from ingestion of food
and on the rate of clearance due to glucose metabolism.
Shown in the figure above are two functions, I(t) and C(t) for the intake and clearance
rates over a period of time. Suppose that at time 0 there is no glucose in the blood. (a)
Express the level of blood glucose as a definite integral. (b) At what time was the intake
rate maximal? (c) At what time was the clearance rate maximal? (d) At what time was
the blood level of glucose maximal?
Solution: (a) It will be
Z
t
I(s) − C(s) ds .
0
(b) About t = 2. (c) About t = 3.4. (d) About t = 2.5.
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