Part 1: Integration, Arc Length, Average Value, Work
Set up the arc-length integral for y=x^{3/2} on 0 \le x \le 4
Formula for arc-length integral: L = \int_a^b \sqrt{1 + (\frac {dy}{dx})²} dx
a = 0
b=4
\frac {dy}{dx} = y^\prime(x)
y^\prime(x) = \frac 32 x^{\frac 12}
L = \int_0^4 \sqrt {1 + (\frac 32 x^{\frac 12})²} dx
L = \int_0^4 \sqrt {1 + \frac 94 x} dx
Set up the arc-length integral for y=e^{2x} on 0\le x \le 1
a = 0
b=1
\frac {dy}{dx} = 2e^{2x}
L = \int_0^1 \sqrt {1 + (2e^{2x})²} dx
L = \int_0^1 \sqrt {1 + 4e^{4x²}} dx
Compute the arc length of the parametric curve
x(t) = t² - 4 \ln t
y(t) = 4\sqrt 2 t
t \epsilon [1,e]
L = \int_a^b \sqrt {(\frac {dx}{dt})² + (\frac {dy}{dt})²} dt
Step 1: Find the derivatives
x^\prime(t) = 2t - \frac 4t
y^\prime(t) = 4\sqrt 2
Step 2: Square the derivatives
(\frac {dx}{dt})² = (2t- \frac 4t)² = 4t² + \frac {16}{t²} - 16
(\frac {dy}{dt})² = (4\sqrt 2)² = 32
L = \int_1^e \sqrt {(2t-\frac 4t)² + (4\sqrt 2)²} dt
L = \int_1^e \sqrt {4t² - 16 + \frac {16}{t²} + 32} dt
L = \int_1^e \sqrt {4t² + \frac {16}{t²} + 16} dt
L = \int_1^e \sqrt {4(t + \frac 2t)²} dt
L = \int_1^e 2 \cdot (t+ \frac 2t) dt = 2\int_1^e t + \frac 2t dt
L = [2\frac {t²}2 + 4\ln|t|]_1^e =[t²+4\ln|t|]_1^e
L = (e² + 4) - (1)
L = e²+3
Compute the arc length of
x(t) = 3t²
y(t) = 2t³
t \epsilon [0,1]
x^\prime(t) = 6t
y^\prime(t) = 6t²
(\frac {dx}{dt})² = 36t²
(\frac {dy}{dt})² = 36t^4
L = \int_0^1 \sqrt {36t² + 36t^4} dt
L = \int_0^1 6t \sqrt{1 + t²} dt
L = 6\int_0^1 t\sqrt {1 + t²} dt
u =1 + t²
du = 2t dt
\frac 12 du = t dt
L = 3 \int_0^1 \sqrt u du
L = 3[\frac 23 \sqrt {u³}]_0^1
L = 3[\frac 23 \sqrt {(1+t²)³}]_0^1
L = 3(\frac 23 \sqrt {8} - \frac 23)
L = 2 \sqrt 8 - 2
L = 4\sqrt 2 - 2
Compute the arc length of:
x(t) = \cos(2t)
y(t) = \sin (2t)
t \epsilon [0,\frac \pi 4]
x^\prime (t) = -2\sin(2t)
y^\prime(t) = 2\cos(2t)
(\frac {dx}{dt})² = (-2\sin (2t))² \Rightarrow 4\sin²(2t)
(\frac {dy}{dt})² = (2\cos(2t))² \Rightarrow 4\cos²(2t)
L = \int_0^{\frac \pi 4} \sqrt {4\sin²(2t) + 4\cos²(2t)} dt
L = 2\int_0^{\frac \pi 4} \sqrt {\sin²(2t) + \cos²(2t)} dt
L= 2 \int_0^{\frac \pi 4} 1 dt
L = 2[x]_0^{\frac \pi 4}
L = 2(\frac \pi 4 - 0)
L = \frac \pi 2
Compute the arc length of
x(t) = 4t²
y(t) = 8t³
t \epsilon [0,1]
x^\prime(t) = 8t
y^\prime (t) = 24t²
(\frac {dx}{dt})² = 64t²
(\frac {dy}{dt})² = 576t^4
L = \int_0^1 \sqrt {64t² + 576t^4} dt
64t² + 576t^4 \Rightarrow 64t²(1 +9t²)
L = 8\int_0^1 t\sqrt {1 + 9t²} dt
u = 1 + 9t²
du = 18t dt
\frac 1{18} du= t dt
L = 8 \cdot \frac 1{18} \int_1^{10} \sqrt{u} du
L = \frac 49 [\frac 23 \sqrt {u³}]_1^{10}
L = \frac 49 (\frac 23 \sqrt {10³} - \frac 23)
Simplified:
L = \frac 8{27}( 10\sqrt {10} - 1)
Set up the integral for computing the length of the curve
x(t) = \cos t
y(t) = \sin (2t)
t \epsilon [0,\pi]
x^\prime (t) = -\sin (t)
y^\prime (t) = 2\cos (t)
(\frac {dx}{dt})² = \sin² (t)
(\frac {dy}{dt})² = 4\cos² (t)
L = \int_0^\pi \sqrt {\sin²(t) + 4\cos²(t)} dt
L = \int_0^\pi \sqrt {\sin²(t) + \cos²(t) + 3\cos²(t)} dt
L = \int_0^\pi \sqrt {1 + 3\cos²(t)} dt
Set up the arc-length integral for:
x(t) = t²
y(t) = \ln t
t \epsilon [1,2]
x^\prime(t) = 2t
y^\prime (t) = \frac 1t
(\frac {dx}{dt})² = 4t²
(\frac {dy}{dt})² = \frac 1{t²}
L = \int_1²\sqrt {4t² + \frac 1{t²}} dt
For the following problems, find the average value of the function on the given interval
Average value equation: \frac 1{b-a} \int_a^b f(x) dx
a) f(x) = 3x² - x + 2 on [0,2]
\frac 1{2-0} \int_0² (3x²-x+2) dx
\frac 12 [x³-\frac {x²}2+2x]_0²
\frac 12 (8 - 2 + 4)
5
b) f(x) = 4\cos x on [0,\pi]
\frac 1{\pi - 0} \int_0^\pi 4\cos x dx
\frac 4\pi [\sin x]_0^\pi
\frac 4\pi (\sin(\pi) - \sin(0))
\frac 4\pi (0 - 0) = 0
c) f(x) = 5x + 1 on [-1,3]
\frac {1}{3 - (-1)} \int_{-1}³(5x+1) dx
\frac 14 [\frac 52 x² + x]_{-1}³
\frac 14 ((\frac 52 (9) + 3)-(\frac 52 - 1))
\frac 14 (24)
6
d) f(x) = e^x on [0,\ln 4]
\frac 1{\ln 4-0}\int_0^{\ln 4} e^x dx
\frac 1{\ln 4} [e^x]_0^{\ln 4}
\frac 1{\ln 4} (4 - 1)
\frac {1}{\ln 4} 3
\frac 3{\ln 4}
e) f(x) = x² + 4 on [-2,2]
\frac 1{2 - (-2)} \int_{-2}² (x² + 4) dx
\frac 14 [\frac 13 x³ + 4x]_{-2}²
\frac 14 ((\frac 83 + 8)-(\frac {-8}3 -8))
\frac 14 (\frac {32}3 + \frac {32}{3})
\frac 14 \frac {64}3
\frac {64}{12} = \frac {16}3
For f(x) = 6 - x² on [0,3]
a) Find the average value
\frac {1}{3}\int_0³ (6-x²) dx
\frac 13 [6x - \frac 13 x ³]_0³
\frac 13 (18 - 9)
\frac 13 \cdot 9
3
b) Find c such that f(x) equals the average value
f(x) = 6 - c² = 3
6-c² = 3
-c² = 3 - 6
c² = 3
c = \sqrt 3
Find the average value of the function on the given interval
a) f(x) = \ln x on [1,e]
\frac 1{e-1} \int_1^e \ln x dx
\frac 1{e-1} [\ln x]_1^e
\frac 1{e-1} (1 - 0)
\frac 1{e-1}1
\frac 1{e-1}
b) f(x) = \frac 1{x+2} on [0,2]
\frac 1{2} \int_0² (\frac 1{x+2}) dx
\frac 12 [\ln|x+2|]_0²
\frac 12 (\ln |4| - \ln |0|)
\frac 12 ()
c) f(x) = \sin²x on [0,\pi]
\frac 1{\pi} \int_0^\pi (\sin² x) dx
\frac 1\pi [\frac {2x-\sin(2x)}4]_0^\pi
\frac 1\pi (\frac {2\pi - \sin (2\pi)}{4}) = \frac 12
Tank temperature T(t) = 20 + 3t, 0\le t\le 4. Find average temperature.
\frac 1{4} \int_0^4 (20 + 3t) dt
\frac 14 [20t + \frac 32 t²]_0^4
\frac 14 (80 + 24)
26
A vertical steel cable (40 m long, 2 N/m) is lifted to the top. Compute the work required.
dW = (weight density)(length element)(distance lifted) = 2dx \cdot (40-x)
W = \int_0^{40} 2(40 - x)dx
W = \int_0^{40} (80 - 2x) dx
W = [80x - x²]_0^{40}
W = (3200 - 1600) = 1600 Joules
Trig Integrals: For the following problems, evaluate the integral
a) \int \cos³ x dx
\int \cos²(x) \cdot \cos(x) dx
\cos²x + \sin²x = 1
\cos²x = 1 - \sin²x
\int \cos x (1-\sin²x) dx
\int \cos x - \cos x \sin²x x dx
\int \cos x dx - \int \sin²x \cos x dx
u = \sin x
du = \cos x dx
\sin x + C - \int u² du
\sin x - \frac 13 u³ + C
\sin x - \frac 13 (\sin x)³ + C
b) \int \sin²(2x) dx
\sin²(\theta) = \frac {1-\cos(2\theta)}{2}
\int \sin²(2x) dx \Rightarrow \int \frac {1-\cos(4x)}{2} dx
\int \frac 12 + \frac {-\cos(4x)}2 dx
\int \frac 12 dx + \int \frac {-\cos(4x)}2 dx
\frac 12x + \frac 12 \int -\cos(4x) dx
u = 4x
du = 4dx
\frac 14 du = dx
\frac x2 + \frac 18 \int -\cos u du
\frac x2 + \frac 18(-\sin(4x)) + C
\frac x2 + \frac {-\sin(4x)}8 + C
c) \int_0^{\frac \pi 4} \sin³x dx
\int_{0}^{\frac \pi 4} \sin³ x dx \Rightarrow \int_0^{\frac \pi 4} \sin²x \sin x dx
\Rightarrow \int_0^{\frac \pi 4} \sin x(1 - \cos² x) dx
\Rightarrow \int_0^{\frac \pi 4} (\sin x - \sin x\cos²x) dx
\Rightarrow \int_0^{\frac \pi 4} \sin x dx + \int_0^{\frac \pi 4} -\sin x\cos²x dx
LHS:
\int_{0}^{\frac \pi 4} \sin x dx \Rightarrow [-\cos x]_0^{\frac \pi 4}
(-\cos (\frac \pi 4) + 1) = (-\frac {\sqrt 2}2 + 1)
RHS:
\int_0^{\frac \pi 4} -\sin x \cos²x dx
u = \cos x
du = -\sin x dx
\int_1^{\frac {\sqrt 2}2} u² du
[\frac 13 u³]_1^{\frac {\sqrt 2}2}
(\frac 13 (\frac {\sqrt 2}2)³ - \frac 13)
\frac {2\sqrt 2}{24} - \frac 13
Total:
-\frac {\sqrt 2}{2} + 1 + \frac {2 \sqrt 2}{24} - \frac 13
-\frac {10 \sqrt 2}{24} + \frac 23
\frac 23 - \frac {5\sqrt 2}{12}
d) \int \sin² x \cos²x dx
\int (\sin x \cos x)² dx
\sin(2x) = 2\sin x \cos x
\sin x \cos x = \frac 12 \sin(2x)
\int (\frac {\sin(2x)}2)² dx
\frac 14 \int \sin²(2x) dx
\sin² \theta = \frac 12 (1-\cos (2\theta))
\frac 14 \int \frac 12(1 - \cos 4x) dx
\frac 18 \int (1 - \cos 4x) dx
\frac 18 (\int 1 dx + \int -\cos 4x dx)
LHS:
\int 1 dx = x + C
RHS:
\int -\cos 4x dx \Rightarrow -\frac 14 \sin (4x) + C
Total:
\frac 18(x - \frac 14 \sin (4x) +C)
\frac x8 - \frac {\sin (4x)}{32} + C
e) \int \tan^5x \sec³x dx
\int \tan^5 x \sec^3 x dx \Rightarrow \int \tan^4x\sec²x \sec x \tan x dx
\Rightarrow \int (\sec² x - 1)² \sec ²x \sec x \tan x dx
u = \sec x
du = \sec x \tan x dx
\Rightarrow \int (u² - 1)² u² du
\Rightarrow \int (u^4 -2u² + 1)u² du
\int (u^6 -2u^4 + u²) du
\Rightarrow \int u^6 du + \int -2u^4 du + \int u² du
1st:
\frac 17 u^7 + C
\frac {\sec^7 x}7 + C
2nd:
-\frac 25 u^5 + C
-\frac {2\sec^5 x}{5} + C
3rd:
\frac 13 u³ + C
\frac {\sec³ x}3 + C
Sum:
\frac {\sec^7 x}{7} - \frac {2\sec^5 x}{5} + \frac {\sec³ x}{3} +C
f) \int \cos^7 (3x) dx
u = 3x
du = 3 dx
\frac 13 du = dx
\frac 13 \int \cos^7(u) du
\frac 13 \int \cos^6 u\cos u du\Rightarrow \frac 13 \int (\cos² u)³ \cos u du
\frac 13 \int (1 - \sin² u)³ \cos u du
v = \sin u
dv = \cos u du
\frac 13 \int (1-v²)³ dv
\frac 13 \int (1+v^4-2x²)(1-v²) dx
\frac 13 \int (1 + v^4 - 2v² - v² - v^6 + 2v^4) dx
\frac 13 \int (1 -v^6 + 3v^4 -3v²) dv
\frac 13 (\int 1 dv + \int -v^6 dv + \int 3v^4 dv + \int -3v² dv)
1st:
v + C
\sin u + C
\sin (3x) + C
2nd:
-\frac {v^7}{7} + C
-\frac {\sin^7u}{7} + C
-\frac {\sin^7(3x)}{7} + C
3rd:
\frac {3v^5}{5} + C
\frac {3\sin(u)^5}{5} + C
\frac {3\sin(3x)^5}{5} + C
4th:
-v³ + C
-\sin³ u + C
-\sin³ (3x) + C
Sum:
\frac 13(\sin(3x) - \frac {\sin^7(3x)}{7} + \frac {3\sin^5(3x)}{5} - \sin³(3x) + C)
\frac {\sin (3x)}{3} - \frac {\sin^7(3x)}{21} + \frac {\sin^5(3x)}{5} - \frac {\sin³(3x)}{3} + C
g) \int \tan^6 x dx
\tan^6x = (\tan² x)³ = (\sec²x - 1)³
(\sec²x - 1)³ = \sec^6 x - 3\sec^4x + 3\sec² x - 1
\int \sec^6 x dx + \int -3\sec^4x dx + \int 3\sec²x dx + \int -1 dx
1st:
\int \sec^6x dx
\int (\sec^4 x)\sec²x dx
\int (\tan²x+1)² \sec²x dx
u=\tan x
du = \sec² x dx
\int (u² + 1)² du
\int (u^4 + 2u² + 1) dx
\frac {u^5}{5} + \frac {2u³}{3} +u
\frac {\tan^5 x}{5} + \frac {2\tan³ x}{3} + \tan x
h) \tan^4 x dx
Trig Substitution:
For the following problems, evaluate the integral.
a) \int \sqrt{9-x²} dx
x = 3\sin(\theta)
dx = 3\cos(\theta) d\theta
\sqrt {9-x²} = 3\cos \theta
\int 3\cos(\theta) \cdot 3\cos(\theta) d\theta
9\int \cos²(\theta) d\theta
9\int \frac {1+\cos(2\theta)}{2} d\theta
\frac 92\int 1 + \cos (2\theta) d\theta
\frac 92 (\theta + \frac 12 \sin (2\theta)) + C
\frac 92 \theta + \frac 94 \sin (2\theta) + C
\theta = \arcsin \frac x3
\sin(2\theta) = \sin \theta \cos \theta + \cos \theta \sin \theta
\sin(2\theta) = 2\sin\theta\cos\theta
2\sin\theta\cos\theta = \frac 23 \frac {\sqrt {9-x²}}{3}
\frac 92 \arcsin \frac x3 + \frac 94 \frac {2x}3 \frac {\sqrt {9-x²}}3 + C
\frac 92 \arcsin \frac x3 + \frac {x\sqrt {9-x²}}2 + C
b) \int \frac {x²}{\sqrt {25-x²}} dx
\sqrt {25 - x²} \Rightarrow
x = 5\sin(\theta)
dx = 5\cos(\theta) d\theta
\sqrt {25 - x²} = 5\cos(\theta)
\int \frac {25\sin²(\theta)}{5\cos(\theta)} 5\cos(\theta) d\theta
\int 25 \sin²(\theta) d\theta
25\int \sin²(\theta) d\theta
\sin²(\theta) = \frac {1-\sin(2\theta)}{2}
25\int \frac {1-\cos(2\theta)}{2} d\theta
\frac {25}2 \int 1 - \cos(2\theta) d\theta
\frac {25}2 (\theta - \frac 12\sin(2\theta)) + C
\frac {25}2\theta - \frac {25}{4}\sin(2\theta) + C
\theta = \arcsin (\frac x5)
\sin (2\theta) = 2\sin\theta\cos\theta = \frac {2x}5 \cdot\frac {\sqrt {25-x²}}{5}
\frac {25}{2} \cdot \arcsin \frac {x}{5} - \frac {25}4 \cdot \frac {2x}5 \cdot \frac {\sqrt {25-x²}}5
\frac {25}{2} \arcsin (\frac x5) - \frac {x}{2} \sqrt {25-x²}
c) \int \sqrt {x² - 16} dx
x = 4\sec(\theta)
dx = 4\sec(\theta)\tan(\theta) d\theta
\sqrt {x² - 16} = 4\tan(\theta)
\int 4\tan (\theta) 4\sec(\theta)\tan(\theta) d\theta
16\int \tan²(\theta)\sec(\theta) d\theta
16\int (\sec²(\theta) -1)\sec (\theta) d\theta
16\int (\sec³(\theta) - \sec(\theta)) d\theta
16\int \sec³(\theta) d\theta + \int -\sec(\theta) d\theta
LHS:
\int \sec³(\theta) d\theta
\int \sec(\theta)\cdot \sec²(\theta) d\theta
\sec \theta \tan \theta - \int \sec \theta \tan² \theta d\theta
= \frac 12 (\sec (\theta) \tan (\theta) + \ln |\sec (\theta) + \tan (\theta)|) + C
RHS:
\int -\sec(\theta) d\theta
-\ln |\sec(\theta) + \tan (\theta) | + C
Combine:
16(\frac 12(\sec(\theta)\tan(\theta) + \ln | \sec(\theta) + \tan(\theta)|) - \ln |\sec(\theta + \tan(\theta)|
8\sec(\theta)\tan(\theta) + 8\ln |\sec(\theta) + \tan(\theta)| - 16\ln |\sec(\theta) + \tan(\theta)| + C
8\sec(\theta)\tan(\theta) - 8\ln|\sec(\theta) + \tan(\theta)| + C
8\frac x4 \frac {\sqrt {x² - 16}}{4}-8\ln |\frac x4 + \frac {\sqrt {x²-16}}{4}| +C
\frac {x\sqrt{x²-16}}2 - 8\ln|\frac {x+\sqrt {x²-16}}{4}| + C
d) \int \frac {1}{x\sqrt {x²-4}} dx
x = 2\sec(\theta)
dx = 2\tan(\theta)\sec(\theta)
\sqrt {x²-4} = 2\tan(\theta)
\int \frac 1{2\sec(\theta)2\tan(\theta)}2\tan(\theta)\sec(\theta) d\theta
\int \frac {1}{2} d\theta
\frac 12 \int d\theta
\frac {\theta}{2}
\frac {\sec^{-1}(\frac x2)}{2}
e) \int \sqrt {x²+36} dx
x=6\tan(\theta)
dx = 6\sec²(\theta)
\sqrt {x²+36} = 6\sec(\theta)
\int 6\sec(\theta) 6\sec²(\theta) d\theta
36\int \sec³(\theta) d\theta
36(\frac 12 (\sec(\theta)\tan(\theta) + \ln|\sec(\theta)+ \tan(\theta)|
18(\frac {\sqrt {x²+36}}6 \cdot \frac x6 + \ln |\frac {\sqrt {x²+36}}{6} + \frac x6|)
\frac {x\sqrt {x²+36}}{2} + 18\ln |\frac {x+\sqrt {x²+36}}{6}|
f) \int_0³ \frac {x}{\sqrt {9 - x²}} dx
x= 3\sin(\theta)
dx = 3\cos (\theta) d\theta
\sqrt {9-x²} = 3\cos(\theta)
\int_0³ \frac {3\sin(\theta)}{3\cos(\theta)} 3\cos(\theta) d\theta
\int_0³ 3\sin(\theta) d\theta
[-3\cos(\theta)]_0³
[-\sqrt {9-x²}]_0³
0 + \sqrt 9 = 3
Partial Fractions:
a) \int \frac {5x-12}{x(x-4)} dx
\frac {5x-12}{x(x-4)} = \frac A{x} + \frac B{(x-4)}
5x-12 = A(x-4) + Bx
5x -12 = Ax - A4 + Bx
5x - 12 = x(A+ B) - A4
A=3
B=2
\int \frac {3}{x} + \frac {2}{(x-4)} d
3\ln |x| + 2\ln|x-4| + C
b) \int \frac {6x-11}{(x-1)²} dx
\frac {6x-11}{(x-1)²} = \frac A{x-1} + \frac B{(x-1)²}
6x-11 = A(x-1) + B
6x-11 = Ax - A + B
A = 6
B= -5
\int \frac{6}{x-1} - \frac {5}{(x-1)²} dx
6\ln |x-1| + \frac 5{x-1} + C
c) \int_1^{\frac 32} \frac {1}{x²-1} dx
d) \int \frac {2x²-25x-33}{(x+1)²(x-5)} dx
e) \int \frac {4x³-3x²+6x-27}{x^4+9x²} dx
f) \int \frac {10}{(x-1)(x²+9)} dx
g) \int \frac {2x}{(x-3)²} dx
Improper integrals: Determine whether the improper integral converges. If it converges, evaluate it.
a) \int_0^{-\infty} e^x dx
[e^x]_0^{-\infty}
e^{-\infty} \Rightarrow 0
e^0 = 1
e^{-\infty} - e^0 = -1
Converges to -1
b) \int_5^{\infty} \frac 1{\sqrt{x-1}} dx
u = x-1
du = 1 dx
\int_4^{\infty} \frac 1{\sqrt u} du = \int_4^{\infty} u^{-\frac 12}
[2\sqrt u]_4^\infty
(\infty) - (2\sqrt 4)
The integral diverges
c) \int_{-\infty}^{\infty} \frac 1{16+x²} dx
\int \frac {1}{a²+x²} = \frac 1a \tan^{-1} (\frac xa)
\int_{-\infty}^{\infty} \frac 1{16+x²} dx \Rightarrow [\frac 14 \tan^{-1} (\frac x4)]_{-\infty}^\infty
\frac 14 \tan^{-1}(\infty) - \frac 14 \tan^{-1}(-\infty)
\frac 14(\frac \pi 2 - (-\frac \pi 2))
\frac 14 (\pi)
\frac \pi 4
d) \int_{1}^{\infty} \frac {1}{x^4 + x²} dx
\frac 1{x²(x² + 1)}
\frac {1}{x²(x²+1)} = \frac A{x²} + \frac B{x²+1}
1 = A(x²+1) + B(x²)
1 = Ax² + A + Bx²
1 = x²(A+B) + A
A = 1
B = -1
\int_1^{\infty} \frac {1}{x²} - \frac 1{x² + 1} dx
[-\frac 1x - \arctan(x)]_1^\infty
-\frac \pi 2 - (-1-\frac \pi 4)
-\frac \pi 2 + 1 + \frac \pi 4
1 - \frac {\pi}4
e) \int_{\frac \pi 4}^{\frac \pi 2} \sec\theta d\theta
\ln |\sec \theta + \tan \theta|_{\frac \pi 4}^{\frac \pi 2}$$
Diverges