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APPM 1350 Exam 3 Solutions Fall 2011
- (^) (a) Use the Fundamental Theorem of Calculus.
y =
∫ (^) x^3
x
t^3 t^2 + 1 dt
=
∫ (^) x 3
0
t^3 t^2 + 1 dt −
∫ (^) x
0
t^3 t^2 + 1 dt
dy dx = d dx
∫ (^) x^3
0
t^3 t^2 + 1 dt − d dx
∫ (^) x
0
t^3 t^2 + 1 dt
= (x^3 )^3 (x^3 )^2 + 1 (3x^2 ) − x^3 x^2 + 1
= 3 x^11 x^6 + 1
x^3 x^2 + 1 (b) Use the product rule. y = x ln(x^2 + 10) y′^ = x ·
x^2 + 10 (2x) + ln(x^2 + 10)
2 x^2 x^2 + 10 + ln(x
(c) Use logarithmic differentiation.
y =
9 x − 1 (2x^3 + 1)^2 (x − 2)^3
ln y = ln (9x^ −^ 1)
1 / 2 (2x^3 + 1)^2 (x − 2)^3 = ln(9x − 1)^1 /^2 − ln(2x^3 + 1)^2 − ln(x − 2)^3 =
ln(9x − 1) − 2 ln(2x^3 + 1) − 3 ln(x − 2) 1 y
dy dx
=^1
9 x − 1
2 x^3 + 1 (6x^2 ) − 3 · 1 x − 2 =
2(9x − 1)
12 x^2 2 x^3 + 1
x − 2
dy dx =
9 x − 1 (2x^3 + 1)^2 (x − 2)^3
2(9x − 1) −^
12 x^2 2 x^3 + 1 −^
x − 2
- (a) Let u = x^3 − 2 , du = 3x^2 dx ⇒ du/3 = x^2 dx. ∫ √^5 x^2 x^3 − 2
dx =^5 3
√du u =
u−^1 /^2 du
=
· 2 u^1 /^2 + C
=
x^3 − 2 + C
(b) The function f (x) = x
(^6) sin x x^2 + 4 is odd because
f (−x) = (−x)^6 sin(−x) (−x)^2 + 4
x^6 (− sin x) x^2 + 4 = − x
(^6) sin x x^2 + 4 = −f (x).
Since
∫ (^) a
−a
f (x) dx = 0 for any odd function f , then
∫ (^3)
− 3
x^6 sin x x^2 + 4 dx = 0.
(c) Let u = 3x − 1 ⇒ x = (u + 1)/ 3 , du = 3 dx ⇒ du/3 = dx. ∫ x
3 x − 1 dx =
u + 1 3
u du 3 =
(u^3 /^2 + u^1 /^2 ) du
=
u^5 /^2 +
u^3 /^2
+ C
(3x − 1)^5 /^2 +
(3x − 1)^3 /^2 + C
(d) Let u = 3x − 2 , du = 3 dx. The u-limits are − 5 to − 2. ∫ (^0)
− 1
3 dx 3 x − 2
− 5
du u = ln |u|
]− 2
− 5
= ln 2 − ln 5 = ln
(e) Let u = sec 2θ, du = 2 sec 2θ tan 2θ dθ. ∫ sec^2 2 θ tan 2θ dθ =^1 2
u du
=
u^2 2
+ C
sec^2 2 θ + C
Alternate solution: Let u = tan 2θ, du = 2 sec^2 2 θ dθ. ∫ sec^2 2 θ tan 2θ dθ =
u du
=
u^2 2
+ C
tan^2 2 θ + C
(Note that tan^2 2 θ + 1 = sec^2 2 θ.)
- (a) True because d dx (x sin x+cos x+C) = x cos x+sin x−sin x = x cos x. (b) True by the Mean Value Theorem for Integrals.
(c) False. Let u = −x, du = −dx. Then
∫ (^) b
a
f (−x) dx =
∫ (^) −b
−a
−f (u) du.
(d) False. (^) nlim→∞
∑^ n
i=
2 i n
n
0
x dx =
, where ∆x =
n and
xi = 2 i n
. The integral 2
0
x dx =
- (a) Rn =
∑^ n i=
i n
n
∑^ n i=
i^2 n^3
= 1 n^3
∑^ n i=
i^2 = 1 n^3 · n(n^ + 1)(2n^ + 1) 6
= (n^ + 1)(2n^ + 1) 6 n^2 = 2 n
(^2) + 3n + 1 6 n^2
(b) (^) lim n→∞ Rn^ =^ nlim→∞
2 n^2 + 3n + 1 6 n^2 (divide by n^2 )
= (^) nlim→∞
2 + (^) n^3 + (^) n^12 6
(c) A definite integral is a limit of Riemann sums:
∫ (^) b
a
f (x) dx =
nlim→∞
∑^ n i=
f (xi) ∆x, where ∆x = b − a n and xi = a + i∆x.
For the given sum, (^) nlim→∞
∑^ n
i=
i n
n
0
x^2 dx , where ∆x =
n and xi = i n
- (a) g is decreasing where g′^ = f < 0 on (2, 4), (6, 8). (b) g has local minimum values at x = 4, 8. (c) g has an absolute minimum value at x = 4. (d) g is concave down where g′′^ = f ′^ < 0 on (2, 3), (5, 7), (9, 10).
- Let f (x) = x^3 − x^2 − x + 1. Then f ′(x) = 3x^2 − 2 x − 1.
x 2 = x 1 − f (x 1 ) f ′(x 1 )
x 3 = x 2 − f (x 2 ) f ′(x 2 ) = 1^ −^
0 =^ undefined Since f (x 2 ) = f (1) = 0, x = 1 is a root of f. Since f ′(x 2 ) = f ′(1) = 0, there is a horizontal tangent at the root.
