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223
CHAPTER 6
Integration
EXERCISE SET 6.
- Endpoints 0,
n
n
n − 1 n , 1; using right endpoints,
An =
[√
n
n
n − 1 n
]
n n 2 5 10 50 100 An 0. 853553 0. 749739 0. 710509 0. 676095 0. 671463
- Endpoints 0,
n
n
n − 1 n , 1; using right endpoints,
An =
[
n n + 1
n n + 2
n n + 3
n 2 n − 1
]
n n 2 5 10 50 100 An 0. 583333 0. 645635 0. 668771 0. 688172 0. 690653
- Endpoints 0, π n
2 π n
(n − 1)π n , π; using right endpoints,
An = [sin(π/n) + sin(2π/n) + · · · + sin(π(n − 1)/n) + sin π] π n n 2 5 10 50 100 An 1. 57080 1. 93376 1. 98352 1. 99935 1. 99984
- Endpoints 0, π 2 n , 2 π 2 n ,... , (n^ −^ 1)π 2 n , π 2 ; using right endpoints,
An = [cos(π/ 2 n) + cos(2π/ 2 n) + · · · + cos((n − 1)π/ 2 n) + co s(π/2)] π 2 n n 2 5 10 50 100 An 0. 555359 0. 834683 0. 919405 0. 984204 0. 992120
- Endpoints 1, n + 1 n
n + 2 n
2 n − 1 n , 2; using right endpoints,
An =
[
n n + 1
n n + 2
n 2 n − 1
]
n n 2 5 10 50 100 An 0. 583333 0. 645635 0. 668771 0. 688172 0. 690653
- Endpoints − π 2
π 2
π n
π 2
2 π n
π 2
(n − 1)π n
π 2 ; using right endpoints,
An =
[
cos
π 2
π n
π 2
2 π n
π 2
(n − 1)π n
( (^) π 2
)]^ π n n 2 5 10 50 100 An 1. 99985 1. 93376 1. 98352 1. 99936 1. 99985
224 Chapter 6
- Endpoints 0,
n
n
n − 1 n , 1; using right endpoints,
An =
n
n
n − 1 n
n
n 2 5 10 50 100 An 0. 433013 0. 659262 0. 726130 0. 774567 0. 780106
- Endpoints − 1 , −1 +
n
n
2(n − 1) n , 1; using right endpoints,
An =
n − 2 n
n − 4 n
n − 2 n
n
n 2 5 10 50 100 An 1 1. 423837 1. 518524 1. 566097 1. 569136
- 3(x − 1) 10. 5(x − 2) 11. x(x + 2) 12.
2 (x^ −^ 1)
2
- (x + 3)(x − 1) 14.
2 x(x^ −^ 2)
- The area in Exercise 13 is always 3 less than the area in Exercise 11. The regions are identical except that the area in Exercise 11 has the extra trapezoid with vertices at (0, 0), (1, 0), (0, 2), (1, 4) (with area 3).
- (a) The region in question is a trapezoid, and the area of a trapezoid is
(h 1 + h 2 )w.
(b) From Part (a), A′(x) =
[f (a) + f (x)] + (x − a)
f ′(x)
=
2 [f^ (a) +^ f^ (x)] + (x^ −^ a)
f (x) − f (a) x − a =^ f^ (x)
- B is alsothe area between the graph of f (x) =
x and the interval [0, 1] on the y−axis, so A + B is the area of the square.
- If the plane is rotated about the line y = x then A becomes B and vice versa.
EXERCISE SET 6.
- (a)
∫ (^) x √ 1 + x^2
dx =
1 + x^2 + C (b)
(x + 1)exdx = xex^ + C
- (a) d dx (sin^ x^ −^ x^ cos^ x^ +^ C) = co sx^ −^ cos^ x^ +^ x^ sin^ x^ =^ x^ sin^ x (b) d dx
x √ 1 − x^2
+ C
1 − x^2 + x^2 /
1 − x^2 1 − x^2
(1 − x^2 )^3 /^2
d dx
[√
x^3 + 5
]
3 x^2 2
x^3 + 5
so
3 x^2 2
x^3 + 5
dx =
x^3 + 5 + C
226 Chapter 6
(sec x tan x + 1)dx = sec x + x + C 25.
sec θ cos θ dθ^ =
sec^2 θ dθ = tan θ + C
sin y dy = − cos y + C 27.
sec x tan x dx = sec x + C
(φ + 2 csc^2 φ)dφ = φ^2 / 2 − 2 co tφ + C 29.
(1 + sin θ)dθ = θ − cos θ + C
2 sin x cos x cos x dx = 2
sin x dx = −2 co sx + C
∫ [
1 − x^2
1 + x^2
]
dx =
sin−^1 x − 3 tan−^1 x + C
∫ [
x
x^2 − 1
1 + x + x^3 1 + x^2
]
dx = 4 sec−^1 x+
x +
x^2 + 1
dx = 4 sec−^1 x+
x^2 +tan−^1 x+C
1 − sin x 1 − sin^2 x
dx =
1 − sin x cos^2 x dx =
sec^2 x − sec x tan x
dx = tan x − sec x + C
1 + co s 2x dx =
2 co s^2 x dx =
sec^2 x dx =
tan x + C
- (a)
2
y
–2 2
x
(b) y
x
4
-1 1
(c) f (x) = x^2 / 2 − 1
- (a) (^5)
6
x
y (^) (b)
1
y
x 1
(c) y = (ex^ + 1)/ 2
- y
x
5
c/ 4 c/ 2
2
y
x 1 2
- f ′(x) = m = − sin x so f (x) =
(− sin x)dx = co sx + C; f (0) = 2 = 1 + C so C = 1, f (x) = co s x + 1
Exercise Set 6.2 227
- f ′(x) = m = (x + 1)^2 , so f (x) =
(x + 1)^2 dx =^1 3 (x + 1)^3 + C;
f (−2) = 8 =
(−2 + 1)^3 + C = −
+ C, = 8 +
, f (x) =
(x + 1)^3 +
- (a) y(x) =
x^1 /^3 dx =^3 4 x^4 /^3 + C, y(1) =^3 4
+ C = 2, C =^5
; y(x) =^3 4 x^4 /^3 +^5 4
(b) y(t) =
(sin t + 1) dt = − cos t + t + C, y
( (^) π 3
2 +^
π 3 +^ C^ = 1/2,^ C^ = 1^ −^
π 3 ; y(t) = − cos t + t + 1 − π 3 (c) y(x) =
(x^1 /^2 + x−^1 /^2 )dx =
3 x
3 / (^2) + 2x 1 / (^2) + C, y(1) = 0 =^8 3 +^ C,^ C^ =^ −^
y(x) =^2 3 x^3 /^2 + 2x^1 /^2 − 8 3
- (a) y(x) =
∫ (^
8 x
− 3
dx = −
16 x
− (^2) + C, y(1) = 0 = − 1 16 +^ C,^ C^ =^
16 ;^ y(x) =^ −^
16 x
− 2 +^1
(b) y(t) =
(sec^2 t − sin t) dt = tan t + co st + C, y( π 4
+ C, C = −
y(t) = tan t + co st −
(c) y(x) =
x^7 /^2 dx =^2 9 x^9 /^2 + C, y(0) = 0 = C, C = 0; y(x) =^2 9 x^9 /^2
- (a) y =
4 ex^ dx = 4ex^ + C, 1 = y(0) = 4 + C, C = − 3 , y = 4ex^ − 3
(b) y(t) =
t−^1 dt = ln |t| + C, y(−1) = C = 5, C = 5; y(t) = ln |t| + 5
- (a) y =
√^3
1 − t^2
dt = 3 sin−^1 t + C, y
= 0 = π + C, C = −π, y = 3 sin−^1 t − π
(b) dy dx = 1^ −^
x^2 + 1 , y^ =
∫ [
x^2 + 1
]
dx = x − 2 tan−^1 x + C,
y(1) = π 2
π 4
- C, C = π − 1 , y = x − 2 tan−^1 x + π − 1
- f ′(x) =
x^3 /^2 + C 1 ; f (x) =
x^5 /^2 + C 1 x + C 2
- f ′(x) = x^2 /2 + sin x + C 1 , use f ′(0) = 2 toget C 1 = 2 so f ′(x) = x^2 /2 + sin x + 2, f (x) = x^3 / 6 − cos x + 2x + C 2 , use f (0) = 1 toget C 2 = 2 so f (x) = x^3 / 6 − cos x + 2x + 2
- dy/dx = 2x + 1, y =
(2x + 1)dx = x^2 + x + C; y = 0 when x = − 3 so( −3)^2 + (−3) + C = 0, C = −6 thus y = x^2 + x − 6
- dy/dx = x^2 , y =
x^2 dx = x^3 /3 + C; y = 2 when x = −1 so (−1)^3 /3 + C = 2, C = 7/ 3
thus y = x^3 /3 + 7/ 3
Exercise Set 6.3 229
- (a) −
u du = − 1 2 u^2 + C = − 1 2 cot^2 x + C
(b)
u^9 du =
u^10 + C =
(1 + sin t)^10 + C
(c)
cos u du =
sin u + C =
sin 2x + C
(d)
sec^2 u du =
tan u + C =
tan x^2 + C
- (a)
(u − 1)^2 u^1 /^2 du =
(u^5 /^2 − 2 u^3 /^2 + u^1 /^2 )du =
u^7 /^2 −
u^5 /^2 +
u^3 /^2 + C
(1 + x)^7 /^2 −
(1 + x)^5 /^2 +
(1 + x)^3 /^2 + C
(b)
csc^2 u du = − cot u + C = − cot(sin x) + C
(c)
sin u du = − cos u + C = − cos(x − π) + C
(d)
du u^2
u
+ C = −
x^5 + 1
+ C
- (a)
u du = ln |u| + C = ln | ln x| + C
(b) −
eu^ du = −
eu^ + C = −
e−^5 x^ + C
(c) −
u du = −
ln |u| + C = −
ln |1 + co s 3θ| + C
(d)
du u = ln^ u^ +^ C^ = ln(1 +^ e
x) + C
- (a) u = x^3 , 1 3
du 1 + u^2
=^1
tan−^1 (x^3 ) + C
(b) u = ln x,
√^1
1 − u^2
du = sin−^1 (ln x) + C
(c) u = 3x,
u
u^2 − 1
du = sec−^1 (3x) + C
(d) u =
x, 2
du 1 + u^2 = 2 tan−^1 u + C = 2 tan−^1 (
x) + C
- u = 2 − x^2 , du = − 2 x dx; −
u^3 du = −u^4 /8 + C = −(2 − x^2 )^4 /8 + C
- u = 3x − 1, du = 3dx;
u^5 du =
18 u
6 + C = 1
18 (3x^ −^ 1)
6 + C
- u = 8x, du = 8dx;^1 8
cos u du =^1 8 sin u + C =^1 8 sin 8x + C
- u = 3x, du = 3dx;^1 3
sin u du = − 1 3 cos u + C = − 1 3 cos 3x + C
230 Chapter 6
- u = 4x, du = 4dx;^1 4
sec u tan u du =^1 4 sec u + C =^1 4 sec 4x + C
- u = 5x, du = 5dx;
sec^2 u du =
tan u + C =
tan 5x + C
- u = 2x, du = 2dx;^1 2
eu^ du =^1 2 eu^ + C =^1 2 e^2 x^ + C
- u = 2x, du = 2dx;
u du =
ln |u| + C =
ln | 2 x| + C
- u = 2x,
√^1
1 − u^2
du =
2 sin
− (^1) (2x) + C
- u = 4x,
1 + u^2 du =
tan−^1 (4x) + C
- u = 7t^2 + 12, du = 14t dt; 1 14
u^1 /^2 du = 1 21 u^3 /^2 + C = 1 21 (7t^2 + 12)^3 /^2 + C
- u = 4 − 5 x^2 , du = − 10 x dx; −
u−^1 /^2 du = −
u^1 /^2 + C = −
4 − 5 x^2 + C
- u = x^3 + 1, du = 3x^2 dx;^1 3
u−^1 /^2 du =^2 3 u^1 /^2 + C =^2 3
x^3 + 1 + C
- u = 1 − 3 x, du = − 3 dx; −
u−^2 du =
u−^1 + C =
(1 − 3 x)−^1 + C
- u = 4x^2 + 1, du = 8x dx;^1 8
u−^3 du = − 1 16 u−^2 + C = − 1 16 (4x^2 + 1)−^2 + C
- u = 3x^2 , du = 6x dx;
cos u du =
sin u + C =
sin(3x^2 ) + C
- u = sin x, du = co sx dx;
eu^ du = eu^ + C = esin^ x^ + C
- u = x^4 , du = 4x^3 dx;
eu^ du =
eu^ + C =
ex 4
- u = − 2 x^3 , du = − 6 x^2 , −
eudu = −
6 e
u (^) + C = − 1 6 e
− 2 x^3 + C
- u = ex^ − e−x, du = (ex^ + e−x)dx,
u du = ln |u| + C = ln
ex^ − e−x
+ C
- u = ex,
1 + u^2 du^ = tan
− (^1) (ex) + C 28. u = t (^2) ,^1 2
u^2 + 1 du^ =
2 tan
− (^1) (t (^2) ) + C
- u = 5/x, du = −(5/x^2 )dx; −
sin u du =
cos u + C =
cos(5/x) + C
- u =
x, du =
x dx;^2
sec^2 u du = 2 tan u + C = 2 tan
x + C
232 Chapter 6
- sec^2 3 θ = tan^2 3 θ + 1, u = 3θ, du = 3dθ ∫ sec^4 3 θ dθ =^1 3
(tan^2 u + 1) sec^2 u du =^1 9 tan^3 u +^1 3 tan u + C =^1 9 tan^3 3 θ +^1 3 tan 3θ + C
1 +^1
t
dt = t + ln |t| + C
- e2 ln^ x^ = eln^ x^2 = x^2 , x > 0, so
e2 ln^ xdx =
x^2 dx =^1 3 x^3 + C
- ln(ex) + ln(e−x) = ln(exe−x) = ln 1 = 0 so
[ln(ex) + ln(e−x)]dx = C
cos x sin x dx; u = sin x, du = co sxdx;
u du = ln |u| + C = ln | sin x| + C
- (a) sin−^1 (x/3) + C (b) (1/
- tan−^1 (x/
5) + C
(c) (1/
π) sec−^1 (x/
π) + C
- (a) u = ex,
4 + u^2 du =
tan−^1 (ex/2) + C
(b) u = 2x,
√^1
9 − u^2
du =
2 sin
− (^1) (2x/3) + C,
(c) u =
5 y,
u
u^2 − 3
du =
sec−^1 (
5 y/
3) + C
- u = a + bx, du = bdx, ∫ (a + bx)ndx =
b
undu = (a + bx)n+ b(n + 1)
+ C
- u = a + bx, du = b dx, dx =
b du 1 b
u^1 /ndu = n b(n + 1) u(n+1)/n^ + C = n b(n + 1) (a + bx)(n+1)/n^ + C
- u = sin(a + bx), du = b cos(a + bx)dx 1 b
undu =
b(n + 1) un+1^ + C =
b(n + 1) sinn+1(a + bx) + C
- (a) with u = sin x, du = co sx dx;
u du =
u^2 + C 1 =
sin^2 x + C 1 ;
with u = co sx, du = − sin x dx; −
u du = −
u^2 + C 2 = −
cos^2 x + C 2
(b) because they differ by a constant: ( 1 2 sin^2 x + C 1
cos^2 x + C 2
(sin^2 x + co s^2 x) + C 1 − C 2 = 1/2 + C 1 − C 2
- (a) First method:
(25x^2 − 10 x + 1)dx =
x^3 − 5 x^2 + x + C 1 ;
second method:^1 5
u^2 du = 1 15 u^3 + C 2 = 1 15 (5x − 1)^3 + C 2
Exercise Set 6.3 233
(b)
(5x − 1)^3 + C 2 =
(125x^3 − 75 x^2 + 15x − 1) + C 2 =
x^3 − 5 x^2 + x −
+ C 2 ;
the answers differ by a constant.
- y(x) =
3 x + 1dx =^2 9 (3x + 1)^3 /^2 + C,
y(1) =
+ C = 5, C =
so y(x) =
(3x + 1)^3 /^2 +
- y(x) =
(6 − 5 sin 2x)dx = 6x +
cos 2x + C,
y(0) =
2 +^ C^ = 3,^ C^ =
2 so^ y(x) = 6x^ +
2 cos 2x^ +
- y(t) =
2 e−t^ dt = − 2 e−t^ + C, y(1) = −
e
+ C = 3 −
e , C = 3; y(t) = − 2 e−t^ + 3
- y =
dx 100 + 4x^2 , u^ =^ x/^5 , dx^ = 5^ du, y =
du 1 + u^2
tan−^1 u + C =
tan−^1
( (^) x 5
π 4
+ C,
C =
π 20 , y^ =^
20 tan
− 1 ( (^) x 5
π 20
- 5
0
-5 5
- y
x
4
-4 4
- f ′(x) = m =
3 x + 1, f (x) =
(3x + 1)^1 /^2 dx =
9 (3x^ + 1)
3 / 2 + C
f (0) = 1 =^2 9
+ C, C =^7
, so f (x) =^2 9 (3x + 1)^3 /^2 +^7 9
- p(t) =
(4 + 0. 15 t)^3 /^2 dt =
(4 + 0. 15 t)^5 /^2 + C; p(0) = 100,000 =
45 /^2 + C =
+ C,
C = 100, 000 −
3 ≈^99 ,^915 , p(t)^ ≈^
3 (4 + 0.^15 t)
5 / (^2) + 99, 915 , p(5) ≈ 8 3 (4.75)
- u = a sin θ, du = a cos θ dθ;
√ du a^2 − u^2
= aθ + C = sin−^1 u a
+ C
- If u > 0 then u = a sec θ, du = a sec θ tan θ dθ,
du u
u^2 − a^2
a θ^ =
a sec
− 1 u a +^ C
Exercise Set 6.4 235
n(n + 1) 2 = 465, n^2 + n − 930 = 0, (n + 31)(n − 30) = 0, n = 30.
- 1 + 2 + 3 +^ · · ·^ +^ n n^2
∑^ n k=
k n^2
n^2
∑^ n k=
k = 1 n^2
n(n + 1) = n^ + 1 2 n ; (^) n→lim+∞^ n^ + 1 2 n
=^1
12 + 2^2 + 3^2 + · · · + n^2 n^3 =
∑^ n
k=
k^2 n^3 =^
n^3
∑^ n
k=
k^2 =
n^3 ·^
6 n(n^ + 1)(2n^ + 1) =
(n + 1)(2n + 1) 6 n^2 ;
n→lim+∞
(n + 1)(2n + 1) 6 n^2 = (^) n→lim+∞
(1 + 1/n)(2 + 1/n) =
∑^ n k=
5 k n^2
n^2
∑^ n k=
k =
n^2
n(n + 1) = 5(n + 1) 2 n ; (^) n→lim+∞ 5(n + 1) 2 n
n∑− 1
k=
2 k^2 n^3
n^3
n∑− 1
k=
k^2 =
n^3
(n − 1)(n)(2n − 1) = (n − 1)(2n − 1) 3 n^2
n→lim+∞
(n − 1)(2n − 1) 3 n^2 = (^) n→lim+∞
(1 − 1 /n)(2 − 1 /n) =
- (a)
∑^5
j=
2 j^ (b)
∑^6
j=
2 j−^1 (c)
∑^7
j=
2 j−^2
- (a)
∑^5
k=
(k + 4)2k+8^ (b)
∑^13
k=
(k − 4)2k
- Endpoints 2, 3 , 4 , 5 , 6; ∆x = 1;
(a) Left endpoints:
∑^4
k=
f (x∗ k)∆x = 7 + 10 + 13 + 16 = 46
(b) Midpoints:
∑^4
k=
f (x∗ k)∆x = 8.5 + 11.5 + 14.5 + 17.5 = 52
(c) Right endpoints:
∑^4
k=
f (x∗ k)∆x = 10 + 13 + 16 + 19 = 58
- Endpoints 1, 3 , 5 , 7 , 9 , ∆x = 2;
(a) Left endpoints:
∑^4
k=
f (x∗ k)∆x =
(b) Midpoints:
∑^4
k=
f (x∗ k)∆x =
(c) Right endpoints:
∑^4
k=
f (x∗ k)∆x =
- Endpoints: 0, π/ 4 , π/ 2 , 3 π/ 4 , π; ∆x = π/ 4
(a) Left endpoints:
∑^4
k=
f (x∗ k)∆x =
(π/4) = π/ 4
236 Chapter 6
(b) Midpoints:
∑^4
k=
f (x∗ k)∆x = [cos(π/8) + cos(3π/8) + cos(5π/8) + cos(7π/8)] (π/4) = [cos(π/8) + cos(3π/8) − cos(3π/8) − cos(π/8)] (π/4) = 0
(c) Right endpoints:
∑^4
k=
f (x∗ k)∆x =
(π/4) = −π/ 4
- Endpoints − 1 , 0 , 1 , 2 , 3; ∆x = 1
(a)
∑^4
k=
f (x∗ k)∆x = −3 + 0 + 1 + 0 = − 2
(b)
∑^4
k=
f (x∗ k)∆x = −
(c)
∑^4
k=
f (x∗ k)∆x = 0 + 1 + 0 − 3 = − 2
- (a) 0.718771403, 0.705803382, 0. (b) 0.668771403, 0.680803382, 0. (c) 0.692835360, 0.693069098, 0.
- (a) 0.761923639, 0.712712753, 0. (b) 0.584145862, 0.623823864, 0. (c) 0.663501867, 0.665867079, 0.
- (a) 4.884074734, 5.115572731, 5. (b) 5.684074734, 5.515572731, 5. (c) 5.34707029, 5.338362719, 5.
- (a) 0.919403170, 0.960215997, 0. (b) 1.076482803, 1.038755813, 1. (c) 1.001028824, 1.000257067, 1.
- ∆x =
n , x∗ k = 1 +
n k; f (x∗ k)∆x =
x∗ k∆x =
n k
n
[
n
n^2 k
]
∑^ n
k=
f (x∗ k)∆x =
[ (^) n ∑
k=
n
∑^ n
k=
n^2 k
]
[
n^2
n(n + 1)
]
[
n + 1 n
]
A = (^) n→lim+∞
[
n
)]
- ∆x =
n , x∗ k = 0 + k
n ; f (x∗ k)∆x = (5 − x∗ k)∆x =
n k
n
n
n^2 k
∑^ n k=
f (x∗ k)∆x =
∑^ n k=
n
n^2
∑^ n k=
k = 25 −
n^2
n(n + 1) = 25 −
n + 1 n
A = (^) n→lim+∞
[
1 +^1
n
)]
=^25
238 Chapter 6
- ∆x =
n , x∗ k = 1 + (k − 1)
n f (x∗ k)∆x =^1 2 x∗ k∆x =^1 2
[
1 + (k − 1)^3 n
]
n
=^1
[
n
]
∑^ n k=
f (x∗ k)∆x =
[ (^) n ∑ k=
n
n^2
∑^ n k=
(k − 1)
]
[
n^2
(n − 1)n
]
n − 1 n
A = (^) n→lim+∞
[
n
)]
- ∆x =
n , x∗ k =
n (k − 1)
f (x∗ k)∆x = (5 − x∗ k)∆x =
[
n (k − 1)
]
n
n
n^2 (k − 1) ∑^ n
k=
f (x∗ k)∆x =
n
∑^ n
k=
n^2
∑^ n
k=
(k − 1) = 25 −
n − 1 n
A = (^) n→lim+∞
[
n
)]
=^25
- ∆x =^3 n , x∗ k = 0 + (k − 1)^3 n ; f (x∗ k)∆x = (9 − 9 (k^ −^ 1)
2 n^2
)^3
n ∑^ n
k=
f (x∗ k)∆x =
∑^ n
k=
[
(k − 1)^2 n^2
]
n =
n
∑^ n
k=
(k − 1)^2 n^2
n^3
∑^ n
k=
k^2 +
n^3
∑^ n
k=
k −
n^2
A = (^) n→lim+∞ = 27 − 27
- ∆x =
n , x∗ k = (k − 1)
n f (x∗ k)∆x =
[
(x∗ k)^2
]
∆x =
[
9(k − 1)^2 n^2
]
n
n
27 k^2 4 n^3
27 k 2 n^3
4 n^3 ∑^ n
k=
f (x∗ k)∆x =
∑^ n
k=
n
4 n^3
∑^ n
k=
k^2 +
2 n^3
∑^ n
k=
k −
4 n^3
∑^ n
k=
4 n^3
n(n + 1)(2n + 1) + 27 2 n^3
n(n + 1) 2
4 n^2
= 12 −
(n + 1)(2n + 1) n^2
4 n
4 n^2
4 n^2 A = (^) n→lim+∞
[
1 +^1
n
2 +^1
n
)]
- ∆x =
n , x∗ k = 2 k − 1 2 n
f (x∗ k)∆x = (2k − 1)^2 (2n)^2
n
k^2 n^3
k n^3
4 n^3 ∑^ n
k=
f (x∗ k)∆x =
n^3
∑^ n
k=
k^2 −
n^3
∑^ n
k=
k +
4 n^3
∑^ n
k=
Using Theorem 6.4.4,
A = (^) n→lim+∞
∑^ n
k=
f (x∗ k)∆x =
Exercise Set 6.4 239
- ∆x =
n , x
∗ k =^ −1 +
2 k − 1 n
f (x∗ k)∆x =
2 k − 1 n
n
8 k^2 n^3
8 k n^3
n^3
n ∑^ n
k=
f (x∗ k)∆x =
n^3
∑^ n
k=
k^2 −
n^3
∑^ n
k=
k +
n^2
A = (^) n→lim+∞
∑^ n k=
f (x∗ k)∆x =
- ∆x =
n , x∗ k = −1 + 2 k n f (x∗ k)∆x =
2 k n
n
n
k n^2 ∑^ n
k=
f (x∗ k)∆x = −2 +
n^2
∑^ n
k=
k = −2 +
n^2
n(n + 1) 2
n
A = (^) n→lim+∞
∑^ n k=
f (x∗ k)∆x = 0
The area below the x-axis cancels the area above the x-axis.
- ∆x =
n , x∗ k = −1 + 3 k n f (x∗ k)∆x =
3 k n
n
n
n^2 k
∑^ n k=
f (x∗ k)∆x = −3 +
n^2
n(n + 1) 2
A = (^) n→lim+∞
∑^ n k=
f (x∗ k)∆x = −3 +^9 2
+ 0 =^3
The area below the x-axis cancels the area above the x-axis that lies tothe right of the line x = 1; the remaining area is a trapezoid of width 1 and heights 1, 2, hence its area is 1 + 2 2
=^3
- ∆x =
n , x∗ k = 2 k n
f (x∗ k) =
[(
2 k n
]
n
8 k^2 n^3
n ∑^ n k=
f (x∗ k)∆x =
n^3
∑^ n k=
k^2 −
n
∑^ n k=
n^3
n(n + 1)(2n + 1) 6
A = (^) n→lim+∞
∑^ n k=
f (x∗ k)∆x =^16 6
− 2 =^2
- ∆x =
n , x∗ k = −1 + 2 k n f (x∗ k)∆x =
−1 +^2 k n
n
n
2 n^3
3 n^4
Exercise Set 6.4 241
- Let A be the area of the region under the curve and above the interval 0 ≤ x ≤ 1 o n thex-axis, and let B be the area of the region between the curve and the interval 0 ≤ y ≤ 1 o n they-axis. Together A and B form the square of side 1, so A + B = 1. But B can alsobe considered as the area between the curve x = y^2 and the interval 0 ≤ y ≤ 1 o n the y-axis. By Exercise 47 above, B =
,so A = 1 −
- If n = 2m then 2m + 2(m − 1) + · · · + 2 · 2 + 2 = 2
∑^ m k=
k = 2 · m(m^ + 1) 2 = m(m + 1) = n
(^2) + 2n 4
if n = 2m + 1 then (2m + 1) + (2m − 1) + · · · + 5 + 3 + 1 =
m∑+
k=
(2k − 1)
m∑+
k=
k −
m∑+
k=
(m + 1)(m + 2) 2 − (m + 1) = (m + 1)^2 = n^2 + 2n + 1 4
∑^30
k=
k(k+20) =
∑^30
k=
k^2 +
∑^30
k=
k =^30 ·^31 ·^61 6
+20^30 ·^31
- both are valid 60. none is valid
∑^ n k=
(ak − bk) = (a 1 − b 1 ) + (a 2 − b 2 ) + · · · + (an − bn)
= (a 1 + a 2 + · · · + an) − (b 1 + b 2 + · · · + bn) =
∑^ n
k=
ak −
∑^ n
k=
bk
∑^ n k=
[
(k + 1)^4 − k^4
]
= (n + 1)^4 − 1 (telescoping sum), expand the
quantity in brackets toget
∑^ n
k=
(4k^3 + 6k^2 + 4k + 1) = (n + 1)^4 − 1 ,
∑^ n k=
k^3 + 6
∑^ n k=
k^2 + 4
∑^ n k=
k +
∑^ n k=
1 = (n + 1)^4 − 1
∑^ n k=
k^3 =
[
(n + 1)^4 − 1 − 6
∑^ n k=
k^2 − 4
∑^ n k=
k −
∑^ n k=
]
[(n + 1)^4 − 1 − n(n + 1)(2n + 1) − 2 n(n + 1) − n]
(n + 1)[(n + 1)^3 − n(2n + 1) − 2 n − 1]
4 (n^ + 1)(n
(^3) + n (^2) ) =^1 4 n
(^2) (n + 1) 2
- (a)
∑^ n k=
1 means add 1 toitself n times, which gives the result.
(b)
n^2
∑^ n
k=
k =
n^2
n(n + 1) 2
2 n , solim n→+∞
n^2
∑^ n
k=
k =
242 Chapter 6
(c) 1 n^3
∑^ n k=
k^2 = 1 n^3
n(n + 1)(2n + 1) 6
=^2
6 n
6 n^2 , solim n→+∞^1 n^3
∑^ n k=
k^2 =^1 3
(d)
n^4
∑^ n
k=
k^3 =
n^4
n(n + 1) 2
2 n
4 n^2 , solim n→+∞
n^4
∑^ n
k=
k^3 =
EXERCISE SET 6.
- (a) (4/3)(1) + (5/2)(1) + (4)(2) = 71/ 6 (b) 2
- (a) (
2 /2)(π/2) + (−1)(3π/4) + (0)(π/2) + (
2 /2)(π/4) = 3(
2 − 2)π/ 8 (b) 3 π/ 4
- (a) (− 9 /4)(1) + (3)(2) + (63/16)(1) + (−5)(3) = − 117 / 16 (b) 3
- (a) (−8)(2) + (0)(1) + (0)(1) + (8)(2) = 0 (b) 2
− 1
x^2 dx 6.
1
x^3 dx
− 3
4 x(1 − 3 x)dx 8.
∫ (^) π/ 2
0
sin^2 x dx
- (a) (^) max ∆limx k →^0
∑^ n
k=
2 x∗ k∆xk; a = 1, b = 2 (b) (^) max ∆limx k →^0
∑^ n
k=
x∗ k x∗ k + 1 ∆xk;^ a^ = 0,^ b^ = 1
- (a) lim max ∆xk → 0
∑^ n k=
x∗ k ∆xk, a = 1, b = 2
(b) (^) max ∆limx k →^0
∑^ n
k=
(1 + co sx∗ k) ∆xk, a = −π/2, b = π/ 2
- (a) A =
2 (3)(3) = 9/^2
3
x
y
A
(b) −A = −
2 (1)(1 + 2) =^ −^3 /^2
-2 -1 x
y
A
(c) −A 1 + A 2 = −
x
y
A 1
A 2
(d) −A 1 + A 2 = 0
x
y
A 1
A 2
