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Math 106a Solutions Exam 2 11/5/
- (a)
sec^3 x tan^3 x dx
Start by splitting off one factor of sec x tan x, use a trigonometric identity to write the tangent function in terms of secant, then substitute u = sec x. ∫ sec^3 x tan^3 x dx =
sec^2 x tan^2 x sec x tan x dx
(sec^4 x − sec^2 x) sec x tan x dx (since tan^2 x = sec^2 x − 1)
(u^4 − u^2 ) du (where u = sec x and du = sec x tan x dx)
= 15 u^5 − 13 u^3 + C = 15 sec^5 x − 13 sec^3 x + C
(b)
θ sec θ tan θ dθ
To start, apply integration by parts with u = θ =⇒ du = dθ dv = sec θ tan θ dθ =⇒ v = sec θ ∫ θ sec θ tan θ dθ = θ sec θ −
sec θ dθ = θ sec θ − ln | sec θ + tan θ| + C
(c)
3 x + 4 (x − 2)(x^2 + 1)
dx
Use partial fractions: ∫ 3 x + 4 (x − 2)(x^2 + 1)
dx =
A
x − 2
Bx + D x^2 + 1
dx
This gives 3x + 4 = A(x^2 + 1) + (Bx + D)(x − 2) therefore
x = 2 =⇒ A = 2
x = 0 =⇒ D = − 1
x = 1 =⇒ B = − 2
∫ 3 x + 4 (x − 2)(x^2 + 1) dx^ =
x − 2 +^
− 2 x − 1 x^2 + 1
dx
x − 2
dx −
2 x x^2 + 1
dx −
x^2 + 1
dx
(let w = x − 2 and dw = dx; & let u = x^2 + 1 and du = 2x dx)
w dw^ −
u du^ −
x^2 + 1 dx = 2 ln |w| − ln |u| − arctan x + C
= 2 ln |x − 2 | − ln |x^2 + 1| − arctan x + C
- The ellipse to the right is given by x^2 + 4y^2 = 1. Solve for y to find the function that represents the top half of the ellipse: x^2 +4y^2 = 1 ⇐⇒ 4 y^2 = 1−x^2 ⇐⇒ y = (^12)
1 − x^2. An integral which represents the top half of the el- lipse is given by 1 2
− 1
1 − x^2 dx
Use trigonometric substitution with x = sin θ for − π 2 ≤ θ ≤ π 2 , then dx = cos θ dθ.
Use sin θ = x to construct a triangle with the opposite side of length x and hypotenuse of length 1. The triangle gives: cos θ =
1 − x^2.
Likewise, our substitution gives θ = arcsin x. To change limits of integration: x = − 1 ⇒ θ = arcsin(−1) = − π 2 x = 1 ⇒ θ = arcsin(1) = π 2
− 1
1 − x^2 dx =
∫ (^) π/ 2
−π/ 2
1 − sin^2 θ cos θ dθ =
∫ (^) π/ 2
−π/ 2
cos^2 θ cos θ dθ (since cos^2 θ = 1 − sin^2 θ)
∫ (^) π/ 2
−π/ 2
cos^2 θ dθ =
∫ (^) π/ 2
−π/ 2
(1 + cos 2θ) dθ (since cos^2 θ = 12 (1 + cos 2θ))
[
θ +
sin 2θ
]π/ 2
−π/ 2
[(
π 2
sin π
π 2
sin(−π)
)]
( (^) π 2
π 2
π 4
- Evaluate
3
(4 − x)^2
dx.
∫ (^6)
3
(4 − x)^2
dx =
3
(4 − x)^2
dx +
4
(4 − x)^2
dx
since 1 (4 − x)^2
is undefined at x = 4
= lim b→ 4 −
∫ (^) b
3
(4 − x)^2
dx + lim a→ 4 +
a
(4 − x)^2
dx (since each integral is improper)
(let u = 4 − x then du = − dx)
= lim b→ 4 −
∫ (^4) −b
1
(−u−^2 ) du + lim a→ 4 +
4 −a
(−u−^2 ) du = lim b→ 4 −
[
u
] 4 −b
1
[
u
]− 2
4 −a
= lim b→ 4 −
4 − b −^1
2 −^
4 − a
But lim b→ 4 −
4 − b
= ∞ and lim a→ 4 +
4 − a
Therefore,
3
(4 − x)^2
dx diverges.
- A 12-inch bar that is clamped at both ends is to be subjected to an increasing amount of stress until it snaps. Let X be the distance from the left end at which the break occurs. Suppose that X has a probability density function given by:
f (x) =
x 24
x 12
if 0 ≤ x ≤ 12
0 otherwise
(a) The probability that the break point occurs between 2 and 4 inches from the left end of the bar is given by: ∫ (^4)
2
x 24
x 12
dx =
2
x −
x^2 12
dx =
[
x^2 2
x^3 36
] 4
2
(b) The expected break point of the bar is given by: ∫ (^) ∞
−∞
xf (x) dx =
−∞
xf (x) dx +
0
xf (x) dx +
12
xf (x) dx
−∞
0 dx +
0
x
x 24
x^2 288
dx +
12
0 dx (refer to definition of f above)
0
x^2 −
x^3 12
dx
[
x^3 3 −^
x^4 48
] 12
0
The expected break point of the bar is 6 inches from the left end.
- Consider
4
3 + sin θ θ^3
dθ.
We have: − 1 ≤ sin θ ≤ 1 for all values of θ
2 ≤ 3 + sin θ ≤ 4 for all values of θ
Thus, 0 ≤
3 + sin θ θ^3
θ^3
for θ > 0
Consider the comparison integral, ∫ (^) ∞
4
θ^3
dθ = lim b→∞
∫ (^) b
4
4 θ−^3 dθ = lim b→∞
[
θ^2
]b
4
= (^) blim→∞
[
b^2
]b
4
Therefore,
4
3 + sin θ θ^3
dθ converges by comparison. In fact,
4
3 + sin θ θ^3
dθ ≤
