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MAT1322 – Notes — By Dr. Hua
- Calculus I Review Contents
- 6.1: Areas Between Curves
- 6.2 Volumes
- 6.3 Volumes by Cylindrical Shells
- 6.4: Work
- 6.5 Average Value of a Function
- 7.8 Improper Integrals
- 8.1 Arc Length
- 8.3: Applications to Physics and Engineering
- 9.1 Modeling with differential equations
- 9.2 Direction Fields and Euler’s method
- 9.3: Separable Equations
- 9.4: Models for Population Growth
- 11.1 Sequences
- 11.2 Series
- 11.3: The Integral Test and Estimates of Sums
- 11.4: The Comparison Tests
- 11.5: Alternating Series and Absolute Convergence
- 11.6: The Ratio and Root Tests
- 11.7: Strategy for Testing Series
- 11.8 Power Series
- 11.9 Representations of Functions as Power Series
- 11.10 Taylor and Maclaurin Series
- 11.11: Applications of Taylor Polynomials
- 12.6: Cylinders and Quadric Surfaces
- 14.1 Functions of Several Variables
- 14.2: Limits and Continuity
- 14.3 Partial Derivatives
- 14.4 Tangent Planes and Linear Approximations
- 14.5 The Chain Rule
14.6 Directional Derivatives and the Gradient Vec-
tor........................................ 79
6.1: Areas Between Curves
Theorem 1.
If f (x) ≥ g(x) for a ≤ x ≤ b, then the area of the region bounded by
y = f (x), y = g(x), x = a, x = b
is
A =
Z (^) b
a
(f (x) − g(x))dx.
Remark. To find the area A of the region enclosed by y = f (x) and y = g(x): Find the intersections, then use the theorem.
Example 1.
Calculate the area of the region bounded by
y = x^2 − 4 x + 7, y = −x^2 + 4x + 1, x = 0, x = 2.
Solution: Step 1. Find intersections: Let (−x^2 +4x+1) = (x^2 − 4 x+7), ⇒ x^2 − 4 x+3 = 0 , ⇒ x = 1, x = 3. Step 2. By using the intersections, the interval (0, 2) is divided into (0, 1) and (1, 2). In (0, 1): x^2 − 4 x + 7 > −x^2 + 4x + 1; In (1, 2): −x^2 + 4x + 1 > x^2 − 4 x + 7. Therefore
A =
Z 1
0
[(x^2 − 4 x + 7) − (−x^2 + 4x + 1)]dx +
Z 2
1
[(−x^2 + 4x + 1) − (x^2 − 4 x + 7)]dx
Z 1
0
(2x^2 − 8 x + 6)dx +
Z 2
1
(− 2 x^2 + 8x − 6)dx.
Example 2.
Find the area of the region between y = x^1 /^2 and y = x^1 /^3 for 0 ≤ x ≤ 1.
Example 3.
Find the area of the region bounded by the parabolas y = 2x − x^2 and y = x^2. Solution: Step 1. Find intersections: 2x − x^2 = x^2 ⇒ x = 0, x = 1. Step 2. For 0 < x < 1, 2x − x^2 > x^2.
A =
Z 1
0
[2x − x^2 − x^2 ]dx = 1 − 23 =^13.
Example 4.
Calculate the area of the region bounded by
y = sin x, y = cos x, x = 0, x = π 2.
Solution: A = 2
Example 5. Find the area of the region bounded by
y = ex, y = 4e−x, y = 1.
Solution: Intersections:
- From ex^ = 4e−x, x = ln 2;
- From ex^ = 1, x = 0;
- From 4e−x^ = 1, x = ln 4.
6.2 Volumes
Volumes of revolution
Case 1: We rotate a region bounded by y = f (x) ≥ 0, x-axis, x = a, x = b around the x-axis (line y = 0). Then the cross section which is perpendicular to x-axis has the area
A(x) = π(f (x))^2 ,
and the volume of the slice is
dV = A(x)dx = π(f (x))^2 dx.
The total volume is
V =
Z (^) b
a
π(f (x))^2 dx.
Example 7. Find the volume of the solid obtained by rotating about the x-axis the region under the curve y =
x from x = 0 to x = 1.
Solution: A(x) = π(√x)^2.
V =
Z 1
0
π(
x)^2 dx =
Z 1
0
πxdx =^1 2
πx^2 |^10 =^1 2
π.
Case 2: We rotate a region bounded by y = f (x), y = g(x), x = a, x = b around x-axis, where f (x) ≥ g(x) ≥ 0. Then the cross section which is perpendicular to x-axis has the area A(x) = πr out^2 − πr^2 in,
where rin = g(x), rout = f (x), dV = A(x)dx,
V =
Z (^) b
a
π[f (x)^2 − g(x)^2 ]dx.
Example 8. √ Find the volume of the solid obtained when the region bounded by y = ex, y = ln x, x = 1 and x = e is rotated around x-axis.
Solution: The cross-sectional area at x is:
A(x) = πr out^2 − πr^2 in = π(e^2 x^ − ln x).
V =
Z (^) e
1
A(x)dx =
Z (^) e
1
π(e^2 x^ − ln x)dx = π(^12 e^2 x^ − x ln x + x)|e 1 = π(^12 e^2 e^ − 12 e^2 − 1).
Case 3: We rotate a region bounded by y = f (x), y = g(x), x = a, x = b around the line y = k (k = 0 ⇐⇒ x-axis), where f (x) ≥ g(x) ≥ 0. Then the cross section which is perpendicular to x-axis has the area
A(x) = πr out^2 − πr^2 in,
Then the cross section which is perpendicular to y-axis has the area
A(y) = πr^2 out − πr in^2 ,
where rin = g(y) − k, rout = f (y) − k, dV = A(y)dy,
V =
Z (^) d
c
[π[f (y) − k]^2 − π[g(y) − k]^2 ]dy.
Example 10. Find the volume of the solid obtained by rotating the region bounded by y = x^3 , y = 8, and x = 0 about the y-axis.
Solution: A(y) = πx^2 = πy^2 /^3.
V =
Z 8
0
πy^2 /^3 dy =^35 πy^5 /^3 |^80 =^1603 π.
6.3 Volumes by Cylindrical Shells
(1) If we rotate a region bounded by y = f (x), y = 0, x = a, x = b about the y-axis, then small shell has the base area π(x + dx)^2 − πx^2 and the height f (x). Thus
dV = {π(x + dx)^2 − πx^2 }f (x) ≈ 2 πxf (x).
V = 2π
Z (^) b
a
xf (x)dx.
(2) If we rotate a region bounded by y = f (x), y = g(x), x = a, x = b (where f (x) ≥ g(x) for x ∈ (a, b)) about the y-axis, then
V = 2π
Z (^) b
a
x[f (x) − g(x)]dx.
(3) If we rotate a region bounded by x = f (y), x = g(y), y = c, y = d (where f (y) ≥ g(y) for y ∈ (c, d)) about the x-axis, then
V = 2π
Z (^) d
c
y[f (y) − g(y)]dy.
Example 11.
Use cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y = x^3 , y = 8, and x = 0 about the y-axis. Solution: When y = 8, x = 2. Thus
V = 2π
Z 2
0
x[8 − x^3 ]dx = 19. 2 π.
Example 12.
Use cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y = 3
x, y = 0, and x = 8 about the x-axis. Solution: When x = 8, y = 2. From y = 3
x we have x = y^3. Thus
V = 2π
Z 2
0
y[8 − y^3 ]dy = 19. 2 π.
6.4: Work
Work done If force is a constant, then: work done= force × distance,
If force is a function f (x), then the work done in moving an object from a to b is:
W =
Z (^) b
a
f (x) dx.
force (in N) = mass (in lb or kg) = mass in kg × acceleration in m/s^2 Work done can have unit ft-lb, or N-m (Joules). 1 ft-lb = 1.36 J, 1lb = 92 ..^82 N = 4.45 N, 1ft=0.305m.
Example 15.
When a particle is located a distance x feet from the origin, a force of 3x^2 + 4x pounds acts on it. How much work is done in moving it from x = 0 to x = 1?
Solution: W =
Z 1
0
(3x^2 + 4x)dx = 3(f t − lb).
Hooke’s Law: The force required to maintain a spring stretched x units beyond its natural length is given by: f (x) = kx, where k > 0 is called the spring constant. The work done to stretch a spring with natural length a meters and force f N from b meters to c meters is:
W =
Z (^) c−a
b−a
kxdx,
where f N is the force required to hold the spring stretched from its natural length to b meters.
Example 16.
A force of 40N is required to hold a spring that has been stretched from its natural length of 10cm to a length of 15cm. How much work is done in stretching the spring from 15cm to 18cm?
Solution: 15-10=5cm=0.05m. f (0.05) = 40, ⇒ 0. 05 k = 40, ⇒ k = 800. Thus f (x) = 800 x.
W =
Z 0. 08
- 05
800 xdx = 1. 56 J.
Example 17.
A cylindrical tank of radius 2 m and height 4 m is half-full of water. Find the work (J) required to pump all of the water to a point that is 2 m above the top of the tank. The density of water is ρ = 1000 kg/m^3 and the acceleration of gravity is g = 9.8 m/s^2. Define clearly all the variables that enter into your solution and provide a drawing which shows their meaning. Solution: We introduce the axis as in the graph.
Take a slice at x with height ∆x. The work done to the slice is:
∆W = ρ∆V gx = ρπ 22 ∆xgx = 4πρgx∆x.
Hence
W =
Z 2+
2+4/ 2
∆W =
Z 6
4
4 πρgxdx = 40πρg = 392000π = 1.. 23 × 106 J.
7.8 Improper Integrals
Type I: Infinite Intervals
The following three are called Type I improper integrals: Z (^) ∞
a
f (x)dx = (^) tlim→∞
Z (^) t
a
f (x)dx, Z (^) b
−∞
f (x)dx = (^) t→−∞lim
Z (^) b
t
f (x)dx, Z (^) ∞
−∞
f (x)dx =
Z ∞
c
f (x)dx +
Z (^) c
−∞
f (x)dx.
Definition:
If the limit is a finite number, we say that the integral is convergent;
If the limit is ±∞, we say that the integral converges to ±∞;
If the limit does not exist at all, we say that the integral is divergent.
Example 20. Evaluate the following improper integrals:
(a)
Z 1
−∞
1 + x^2
dx
(b)
Z ∞
−∞
e−|x|dx.
Solution: (a). Z (^1)
−∞
1 + x^2 dx^ = lim^ t→∞
Z (^) t
1
1 + x^2 dx^ = lim^ t→∞(arctan^ t^ −^ arctan 1) =^
π 2 −^
π 4 =
3 π
(b). (^) Z ∞ −∞
e−|x|dx =
Z 0
−∞
e−|x|dx +
Z ∞
0
e−|x|dx =
Z 0
−∞
exdx +
Z ∞
0
e−xdx
= (^) t→−∞lim
Z 0
t
exdx + (^) t→−∞lim
Z (^) t
0
e−xdx = (^) t→−∞lim (1 − et) + (^) t→−∞lim(e−t^ − 1) = 2.
Example 21. Determine if the integral
I =
Z ∞
2
xe−xdx
is convergent or divergent and evaluate if it is convergent.
Solution: I = lim t→∞
Z (^) t
2
xe−x^ dx (integration by parts: let u = x and dv = e−xdx)
= lim t→∞
x(−e−x)|t 2 −
Z (^) t
2
−e−x^ dx
= lim t→∞(−xe−x^ − e−x)|t 2 = lim t→∞ −(x + 1)e−x|t 2 = lim t→∞(−(t + 1)e−t^ + 3e−^2 ) = 3e−^2 (where limt→∞(t + 1)e−t^ = 0 by L’Hospital’s Rule).
Type 2: Discontinuous Integrands
If f (x) is continuous on [a, b), but discontinous at x = b, then
Z (^) b
a
f (x)dx = lim t→b−
Z (^) t
a
f (x)dx;
If f (x) is continuous on (a, b], but discontinous at x = a, then Z (^) b
a
f (x)dx = lim t→a+
Z (^) b
t
f (x)dx;
If f (x) is discontinuous at c: a < c < b, then Z (^) b
a
f (x)dx =
Z (^) c
a
f (x)dx +
Z (^) b
c
f (x)dx.
Example 22. Determine if the following integrals are convergent or divergent and evaluate if it is convergent.
(a)
Z 3
2
√^1
3 − x
dx,
(b)
Z 2
0
x − 1 dx,
(c)
Z (^) e
0
ln xdx.
√^1
x^3 + 1
≤ (^) x^13 / 2.
Example 26. (^) Z (^) ∞
8
1 + √x x − 6 dx^ =^ divergent. ∵ 1 +
x x − 6 ≥^
√^1
x.
Example 27. Determine whether the integral Z (^) ∞
2
cos^4 x ex^ + sin^2 x + 1
dx
is convergent or divergent.
Solution: Since cos^4 x ex^ + sin^2 x + 1
≤ (^) e^1 x = e−x, Z (^) ∞
2
cos^4 x ex^ + sin^2 x + 1dx^ ≤
Z ∞
2
e−xdx = e−^2.
By Comparison Theorem, the original integral is convergent.
Example 28. Determine whether the integral Z (^1)
0
x^1.^9 sin^2 x
dx
is convergent or divergent.
Solution: Let y = 1/x. Then Z (^1)
0
x^1.^9 sin^2 x
dx =
Z ∞
1
y^0.^1 sin2 1 y
dy ≥
Z ∞
1
y^0.^1
dy = ∞.
8.1 Arc Length
The length of the arc given by the parametric equations x = x(t), y = y(t) (a ≤ t ≤ b) is
L =
Z (^) b
a
s dx dt
dy dt
dt.
There are two special cases t = x or t = y:
L =
Z (^) b
a
s 1 +
dy dx
dx, L =
Z (^) d
c
s dx dy
Example 29.
Calculate the length of the arc y = 12 (ex^ + e−x) between x = 0 and x = 2. Solution:
L =
Z 2
0
p 1 + [f ′(x)]^2 dx =
Z 2
0
r 1 + [
2 (e
x (^) − e−x)] (^2) dx =
Z 2
0
r 1 +
4 (e
2 x (^) − 2 + e− 2 x)dx
Z 2
0
r 1 4 (e
2 x (^) + 2 + e− 2 x)dx =
Z 2
0
2 (e
x (^) + e−x)dx =^1 2 (e
x (^) − e−x)| (^20)
=^1
(e^2 − e−^2 )
Example 30.
Calculate the length of the arc
x(t) = t^2 + 10, y(t) =^2 3
t^3 , 0 ≤ t ≤ 3.
Solution: x′(t) = 2t, y′(t) = 2t^2. Hence
L =
Z 3
0
p x′(t)^2 + y′(t)^2 dt =
Z 3
0
p (2t)^2 + (2t^2 )^2 dt
Z 3
0
2 t
1 + t^2 dt =
Z 3
0
1 + t^2 d(1 + t^2 )
=^23 (1 + t^2 )^3 /^2 |^30 =^23 [(10)^3 /^2 − 1] = 20. 41.
