(c) (6 points) lim x→0 tan22x 3x2

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You need to show your work to obtain credits 1. State whether the limit exists; evaluate the limit if it does exist.

(a) (6 points) lim

x→2⁺

x− 2

|x − 2|. (b) (6 points) lim

x→0

1

x−1 − x x
.

(c) (6 points) lim

x→0

tan²2x 3x² . Solution:

(a) lim

x→2⁺

x− 2

|x − 2| = lim

x→2⁺

x− 2 x− 2 = 1.

(b) lim

x→0

1

x−1 − x

x
= lim

x→0

1 − 1 + x x = 1.

(c) By using the formula lim

θ →0

sin θ

θ = 1, we have

x→0lim

tan²2x 3x² = lim

x→0

sin 2x 2x

2 4

3 cos²2x
= 4 3.

2. Find dy

dx for the following (a) (6 points) y = 1 + 2x + x²

x³− 1 . (b) (6 points) y = tan√

2x + 1.

(c) (6 points) y = lnx²
3

.

Solution:

(a) By the Product rule, we get dy

dx =2 + 2x

x³− 1−(1 + 2x + x²)3x²

(x³− 1)² = (2 + 2x)(x³− 1) − (1 + 2x + x²)3x² (x³− 1)²

= −2 − 2x − 3x²− 4x³− x⁴ (x³− 1)² . (b) By the Chain rule, we get

dy

dx = sec²√

2x + 1 d dx

√

2x + 1 = sec²√ 2x + 1

√2x + 1 .

(c) By the Chain rule, we get dy

dx = 3 lnx²
2 d

dxlnx²= 3 lnx²
22x

x² =6 lnx²
2

x .

3. (a) (6 points) Let y = Z _x

0

dt

1 + t⁴. Find dy dx.

(b) (6 points) Let 2x³+ 3x cos y = 2xy. Use implicit differentiation to express dy

dx in terms of x and y.

Solution:

(a) dy

dx = 1 1 + x⁴.

(b) By differentiating both sides with respect to x, we get 6x²+ 3 cos y − 3x sin ydy

dx = 2y + 2xdy dx. Thus, we have

dy

dx =6x²+ 3 cos y − 2y 2x + 3x sin y .

4. (8 points) Let C₁and C₂be curves in the xy-plane defined in polar coordinates by C₁: r = 2 cos θ , C₂: r = 2√

3 sin θ . Sketch C₁and C₂and find the xy-coordinates of the points of intersections.

Solution: In terms of the xy-coordinates,

C₁: x²+ y²= 2x ⇔ (x − 1)²+ y²= 1.

C₂: x²+ y²= 2√

3y ⇔ x²+ (y −√

3)²= 3.

C₁and C₂intersect when

r= 0 ⇔ (x, y) = (0, 0) or

2 cos θ = 2√

3 sin θ ⇔ tan θ = 1

√

3 ⇔ cos θ =

√3

2 , sin θ = 1

2, r =√

3 ⇔ (x, y) = (3 2,

√3 2 ).

5. (a) (6 points) Use the substitution s = t^2/3to find

Z t^2/3− 1
2

t^1/3 dt. (b) (6 points) Use the substitution y = 2 − x to find

Z x√

2 − x dx.

(c) (6 points) Use the substitution u = sin x to find

Z cos x 1 + sin²xdx.

Solution:

(a) By using the substitution s = t^2/3, we have ds dt = 2

3t^1/3 and Z t^2/3− 1
2

t^1/3 dt = 3 2 Z

s− 1
2

ds= 1

2 s− 1
3

+C =1

2 t^2/3− 1
3

+C.

(b) By using the substitution y = 2 − x, we have dy

dx= −1 and Z

x√

2 − x dx = Z

(y − 2)√ y dy=

Z

(y^3/2− 2y^1/2) dy

= 2

5y^5/2−4

3y^3/2+C = 2

5(2 − x)^5/2−4

3(2 − x)^3/2+C.

(c) By using the substitution u = sin x, we have du

dx = cos x and Z cos x

1 + sin²xdx=

Z 1

1 + u²du= tan⁻¹u+C = tan⁻¹sin x +C.

6. (a) (6 points) Use partial fractions to find

Z x− 1 x(x + 1)dx.

(b) (6 points) Use the integration by parts formula to find Z

ln(x + 1) dx.

Solution:

(a) By setting x− 1 x(x + 1)= A

x + B

x+ 1 = (A + B)x + A

x(x + 1) , we get A = −1, B = 2.

Z x− 1

x(x + 1)dx= Z −1

x + 2

x+ 1dx= −ln|x| + 2ln|x + 1| +C, or ln(x + 1)²

|x| +C.

(b) By using the integration by parts formula, we get Z

ln(x + 1) dx = xln(x + 1) − Z x

x+ 1dx= xln(x + 1) − Z

1 − 1 x+ 1
dx

= xln(x + 1) − x + ln(x + 1) +C.

7. (a) (8 points) Find the area inside one petal of r = 2 cos 4θ for θ ∈ [−π 8,π

8].

(b) (6 points) Find the length of the curve y = x^3/2+ 2 from x = 0 to x =5 9. Solution:

(a) The area inside one petal of r = 2 cos 4θ is equal to 1

2 Z _{π /8}

−π/8

r²dθ = Z _{π /8}

0

r²dθ = 4 Z _{π /8}

0

cos²4θ dθ = 2 Z _{π /8}

0 (1 + cos 8θ ) dθ

= 2 θ +sin 8θ

8
|^{π /8}₀ = π 4.

(b) The length of the curve y = x^3/2+ 2 from x = 0 to x = 5

9 is equal to Z _5/9

0

r

1 + dy dx

2

dx= Z _5/9

0

r 1 +9x

4 dx= 2 3·4

9 1 +9x 4

3/2

|^5/9₀

= 8 27(9

4)^3/2− 1 = 8 27

 27

8 − 1 = 19 27.