Calculus 1, Quiz 4
(1) (28 Points) Evaluate
2 sin2 x + sin x − 1
.
(a) limπ
x→ 6 2 sin2 x − 3 sin x + 1
Solution:
For x 6=
π
, we have
6
(2 sin x − 1)(sin x + 1)
2 sin2 x + sin x − 1
=
2
(2
sin x − 1)(sin x − 1)
2 sin x − 3 sin x + 1
sin x + 1
=
.
sin x − 1
Hence
limπ
x→ 6
(b) lim
x→0
sin π6 + 1
2 sin2 x + sin x − 1
sin x + 1
=
=
=
lim
sin π6 − 1
2 sin2 x − 3 sin x + 1 x→ π6 sin x − 1
1
2
1
2
+1
= −3.
−1
1 − cos x
.
x2
Solution:
For x 6= 0,
2 sin2 x2
1 − cos x
=
x2
x
2
1 sin x2
=
.
x
2
2
sin θ
x
. Since lim
, we find
θ→0 θ
2
2
1 − cos x
1 sin x2
1
1
lim
=
lim
= · 12 = .
x
x→0
x→0 2
x2
2
2
2
Here we use 1 − cos x = 2 sin2
(c) lim
x→0
tan 3x
.
sin 8x
Solution:
For x 6= 0, we write
8x
sin 3x
1
3
tan 3x
=
·
·
· .
sin 8x
sin 8x
3x
cos 3x 8
sin θ
Since lim
= cos θ = 1, we find
x→0 θ
tan 3x
8x
sin 3x
1
3
3
3
lim
= lim
·
·
·
= ·1·1·1= .
x→0 sin 8x
x→0 sin 8x
3x
cos 3x 8
8
8
1
2
x + x2 + · · · + xn − n
.
x→1
x−1
(d) lim
Solution:
Write xk − 1 = (x − 1)(xk−1 + · · · + x + 1). Hence
2
n
x + x + ··· + x − n =
=
=
n
X
!
x
k
−n
k=1
n
X
(xk − 1)
k=1
n
X
(x − 1)(xk−1 + · · · + x + 1)
k=1
= (x − 1)
n
X
(xk−1 + · · · + x + 1).
k=1
For x 6= 1,
n
x + x2 + · · · + xn − n X k−1
=
(x
+ · · · + x + 1).
x−1
k=1
By properties of limits and the continuity of polynomials, we have
n
n(n + 1)
x + x2 + · · · + xn − n X
=
k=
.
lim
x→1
x−1
2
k=1
√
(e) lim
x→0
1 + tan x −
x3
√
1 + sin x
.
Solution:
Using the formula A2 − B 2 = (A + B)(A − B), we write
√
√
1 + tan x −
√
tan x − sin x
√
1 + tan x + 1 + sin x
sin x
1
√
− sin x
=√
1 + tan x + 1 + sin x cos x
1
sin x
√
=√
·
(1 − cos x).
1 + tan x + 1 + sin x cos x
1 + sin x = √
Hence we know
√
1 + tan x − 1 + sin x
1
sin x
1
1 − cos x
√
=√
·
·
·
.
x3
x
cos x
x2
1 + tan x + 1 + sin x
3
1
1
sin x
1
√
= , and lim
= 1 and lim
= 1 and
x→0
x→0
2
x
cos
x
1 + tan x + 1 + sin x
1
1 − cos x
= , we obtain
lim
x→0
x2
2
√
√
1
1 + tan x − 1 + sin x
1
1
lim
= ·1·1· = .
3
x→0
x
2
2
4
p
p
3
(f) lim ( x3 + 3x2 − x2 − 2x).
Note that lim √
x→0
x→∞
Solution:
x We use the formula A6 − B 6 = (A − B)(A5 + A4 B + A3 B 2 + A2 B 3 + AB 4 + B 5 ).
Write
p
p
3
x3 + 3x2 − x2 − 2x = ((x3 + 3x2 )2 − (x2 − 2x)3 )
4
5
1
· {(x3 + 3x2 ) 3 + (x3 + 3x2 ) 3 (x2 − 2x) 2
4
2
1
4
2
3
+ (x3 + 3x2 ) 3 (x2 − 2x) 2 + (x3 + 3x2 ) 3 (x2 − 2x) 2
5
+ (x3 + 3x2 ) 3 (x2 − 2x) 2 + (x2 − 2x) 2 }−1
8
3
= (12 − + 2 )
x x
3 5
3 4
2 1
· {(1 + ) 3 + (1 + ) 3 (1 − ) 2
x
x
x
3 3
2 2
3 2
2 3
+ (1 + ) 3 (1 − ) 2 + (1 + ) 3 (1 − ) 2
x
x
x
x
3 1
2 4
2 5 −1
+ (1 + ) 3 (1 − ) 2 + (1 − ) 2 } .
x
x
x
Hence
p
3
x3 + 3x2 −
x→∞
3π
(g) limπ cos 2x + sin(
+ x) .
x→ 2
2
lim (
p
12
x2 − 2x) =
= 2.
6
Solution:
3π
3π
limπ cos 2x + sin(
+ x) = cos limπ 2x + limπ sin(
+ x)
x→ 2
x→ 2
x→ 2
2
2
π
= cos(2 · + sin 2π)
2
= cos π = −1.
(2) (12 Points)
Evaluate
Z √23 p
(a)
1 − x2 dx.
1
√
2
4
The
Solution:
integral is the area of the region DEIH. We write
Z √23 p
Z √23 p
Z √1 p
2
2
2
1 − x dx =
1 − x dx −
1 − x2 dx.
1
√
2
0
0
√
Z
3
2
We know
p
1 − x2 dx equals the area of the region AEIC and
0
equals the area of the region ADHC.
Since the area of the region EBI is given by
√
√
π
1
3 1
π
3
− ·
· (area of ∆AEI) =
−
,
12 2 2 2
12
8
the area of the region AEIH is given by
√ !
√
π
3
π
3
π
−
−
= +
.
4
12
8
6
8
In other words,
√
π
3
+
.
6
8
0
Similarly, we know the area of the region DBH is given by
π 1 1
1
π 1
− · √ · √ (area of ∆ADH) = − .
8
2
8
4
2
2
√
Z
3
2
p
1 − x2 dx =
Hence the area of the region ADHC is given by
π
π 1
π 1
−
−
= + .
4
8
4
8
4
Z
0
1
√
2
p
1 − x2 dx
5
In other words,
Z
0
1
√
2
p
π 1
1 − x2 dx = + .
8
4
We conclude that
Z
√
3
2
p
1 − x2 dx =
1
√
2
Z
√ ! √
3
3−2
π
π 1
π
+
−
+
=
+
.
6
8
8
4
24
8
6
(b)
[x]dx. Here [x] is the Gauss-symbol of x.
3
Solution:
The integral of f (x) over [3, 6] is the sum of all areas of the region I, II, II. Hence
Z
6
[x]dx = 1 · 3 + 1 · 4 + 1 · 5 = 12.
3
Z
2
(c)
f (x)dx. Here
−1


2x + 2

−4x − 1
f (x) =
x − 1



1
Solution:
if
if
if
if
−1 ≤ x ≤ −1/2,
−1/2 < x ≤ 0,
0 < x ≤ 1,
1 < x ≤ 2.
6
Z
2
The integral
f (x)dx equals the sum of areas of I and III minus the area of II, i.e.
−1
Z
2
f (x)dx = A(I) − A(II) + A(III).
−1
The area of I, II, II are given by
3
1 3
A(I) = · · 1 = (area of ∆ABH)
2 4
8
1 5
5
A(II) = · · 1 = (area of ∆CDH)
2 4
8
A(III) = 1 · 1(area of the square DGF E).
Hence
Z
2
f (x)dx =
−1
3
3 5
− +1= .
8 8
4