Final Review - Chapter 5: The Integral
5.7. EX.1: Evaluate. (3/4)∫0 dx/√(9-16x^2)
(3/4)∫0 dx/√(9-16x^2) First note that you can rewrite the integrand to be: √(9-16x^2) = √9(1-16x^2/9) = 3√(1-(4x/3)^2) Thus: (3/4)∫0 dx/√(9-16x^2) = (3/4)∫0 dx/3√(1-(4x/3)^2) u = 4x/3 du = (4/3)dx (3/4)du = dx u(3/4) = 4(3/4)/3 = 3/3 = 1 u(0) = 0 = 1∫0 (3/4)du/3√(1-u^2) = (3/4) * (1/3) 1∫0 du/√(1-u^2) = (1/4) * 1∫0 du/√(1-u^2) = (1/4) * (arcsin(u) 1|0) = (1/4) * (arcsin(1) - arcsin(0)) = (1/4) * (π/2 - 0) = (1/4) * (π/2) = π/8
4.9. Initial Condition
1. Given f(x), Find the general anti-derivative (integral) F(x). 2. Given the initial condition F(a) = b, set F(a) + C = b to get the constant.
5.2. Properties of the Definite Integral
1. b∫a C dx = C(b-a) 2. b∫a [f(x)+g(x)] dx = b∫a f(x)dx + b∫a g(x)dx b∫a [f(x)-g(x)] dx = b∫a f(x)dx - b∫a g(x)dx 3. b∫a Cf(x) dx = C * b∫a f(x)dx 4. a∫a f(x)dx = 0 5. b∫a f(x)dx = - a∫b f(x)dx 6. b∫a f(x)dx = c∫a f(x)dx + b∫c f(x)dx
5.1. Rectangle Approximation
1. Δx = (b - a)/N 2. Add up all right/left endpoints or midpoints (divide the interval into N subintervals). a. Right: Skip a, up to b. b. Left: Start at a, skip b. c. Start at a + 1/2, up to b - 1/2. 3. Multiple this sum by Δx.
5.3. Basic Antiderivative Formulas
1. ∫ x^ndx = x^(n+1)/(n+1) + C for n =/= 1 2. ∫ e^xdx = e^x + C 3. ∫ dx/x = ln|x| + C 4. ∫ sinxdx = -cosx + C 5. ∫ cosxdx = sinx + C 6. ∫ sec^2xdx = tanx + C 7. ∫ csc^2xdx = -cotx + C 8. ∫ secxtanxdx = secx + C 9. ∫ cscxcotxdx = -cscx + C
5.2. EX. 1: Use properties of the integral to calculate the integrals: 1∫-3 (7x^2 + x + 1)dx
1∫-3 (7x^2 + x + 1)dx = 7 * 0∫-3 (x^2)dx + 1∫-3 xdx + 1∫-3 1dx = 7(-3∫0 x^2dx + 1∫0 x^2dx) + 0∫-3 xdx + 1∫0 xdx + 0∫-3 1dx + 1∫0 1dx = 7( -(1/3)(-3)^3 + (1/3)(1)^3) ) - (1/2)(-3)^2 + (1/2)(1)^2 + (-(-3)) + 1 = 7(9 + 1/3) - 9/2 + 1/2 + 3 +1 = 392/6 - 27/6 + 3/6 + 18/6 + 6/6 = 392/6 = 196/3
5.6. EX. 2: Evaluate the definite integral. 1∫0 (x+1)(x^2+2x)^5dx
1∫0 (x+1)(x^2+2x)^5dx u = x^2+2x u(0) = 0^2 + 2(0) = 0 u(1) = 1^2 + 2(1) = 3 du = (2x+2)dx du = 2(x+1)dx (1/2)du = (x+1)dx = 3∫0 u^5 * (1/2)du = (1/2) * 3∫0 u^5du = (1/2)[(1/6 * u^6) 3|0] = (1/2)(1/6 * 3^6) = (1/2)(1/6 * 729) = (1/2)(729/6) = 729/12 = 243/4
5.3. EX. 1: Calculate 3∫-2 f(x)dx, where: f(x) = {12-x^2 for x<=2, x^3 for x>2}
3∫-2 f(x)dx = 2∫-2 f(x)dx + 3∫2 f(x)dx = 2∫-2 (12-x^2) dx + 3∫2 (x^3) dx = (12x - x^3/3) 2|-2 + (x^4/4) 3|2 = (24 - 8/3 + 24 - 8/3) + (81/4 - 16/4) = 48 - 16/3 + 65/4 = 128/3 + 65/4 = 707/12
5.1. Summation Laws
Addition (- for Subtraction): (Separate each term into multiple sums) (N) Σ (j=m) {a + b} = (N) Σ (j=m) a + (N) Σ (j=m) b Coefficient: (Pull out constant to front) Σ (j=m) {ka}= k * Σ (j=m) a Constant: Σ (j=m) {k} = N * k
5.3. Fundamental Theorem of Calculus Part I* (*Most often referred to as FTC II)
Assume that f (x) is continuous on [a, b]. If F (x) is an antiderivative of f (x) on [a, b], then: b∫a f(x)dx = F(b) - F(a) = F(x) b|a where F'(x) = f(x)
5.4. Fundamental Theorem of Calculus Part II* (*Most often referred to as FTC I)
Assume that f (x) is continuous on an open interval I and let a I. Then the area function: A(x) = x∫a f(t)dt And is an antiderivative of f (x) on I; that is: A'(x) = f(x) = d/dx x∫a f(t)dt Special Case/Initial Condition : A(a) = a∫a f(t)dt = 0
5.5. EX. 2: A particle moves in a straight line with the velocity (in m/s) v (t) = 12 − 4t. Find the displacement and distance traveled over the time interval [0, 5].
Displacement = 5∫0 v(t)dt = 5∫0 (12-4t)dt = (12t-2t^2) 5|0 = 12(5) - 2(25) = 60 - 50 = 10 meters For distance traveled, we must find the time t when the velocity changes direction (i.e. it crosses the x-axis): 12 - 4t = 0 12 = 4t 3 = t Find out which portion of the interval the velocity is negative: v(0) = 12 - 4(0) = 12 -> positive v(5) = 12 - 4(5) = -8 -> negative Therefore: Distance Traveled = area above - area below = 5∫0 |v(t)|dt = 5∫0 |12-4t|dt = 3∫0 (12-4t)dt - 5∫3 (12-4t)dt = (12t-2t^2) 3|0 - (12t-2t^2) 5|3 = [(36 - 18) - 0] - [(60 -50) - (36 - 18)] = 18 - (10 - 18) = 18 + 8 = 26 meters
5.5. Displacement vs Distance Traveled
Displacement over [t1, t2] = t2∫t1 v(t)dt = s(t2) - s(t1) Total distance traveled over [t1, t2] = t2∫t1 |v(t)|dt = s(t2) - s(t1) If you travel 10 km and return to your starting point, your displacement is zero but your distance traveled is 20 km.
5.1. Distance
Distance traveled is equal to the area under the graph. It is approximated by the sum of the areas of the rectangles. d(t) = v*(t1 - t2) for constant velocity d(t) = ∫v(t)dt for velocity that changes over time
5.4. EX. 2: Calculate the derivative. d/dx x^2∫sqrt(x) tan(t)dt
F(x) = x^2∫sqrt(x) tan(t)dt = F(x^2) - F(sqrt(x)) F'(x) = d/dx x^2∫sqrt(x) tan(t)dt = F'(x^2) - F'(sqrt(x)) F'(x) = tan(x) x^2|sqrt(x) = 2x*tan(x^2) - tan(sqrt(x))/2sqrt(x) **When taking the antiderivative of a function, you must use the chain rule.
5.3. EX. 3: Calculate F(4) given that F(1) = 3 and F'(x) = x^2.
FTC 1: b∫a f(x)dx = F(b) - F(a) where F'(x) = f(x) 4∫1 x^2dx = F(4) - F(1) (1/3x^3) 4|1 = F(4) - 3 (1/3)(4^3) - (1/3)(1^3) = F(4) - 3 64/3 - 1/3 = F(4) - 3 63/3 + 3 = F(4) 21 + 3 = F(4) 24 = F(4)
5.1. Power Sums
First power: (N) Σ (j=1) j = N(N+1) / 2 = N^2/2 + N/2 Second power: (N) Σ (j=1) j^2 = N(N+1)(2N+1) / 6 = N^3/3 + N^2/2 + N/6 Third power: (N) Σ (j=1) j^3 = N^2(N+1)^2 / 4 = N^4/4 + N^3/2 + N^2/4
5.4. EX. 1: Let G(x) = x∫1 (t^2-2)dt. Calculate G(1), G'(1) and G'(2). Then find a formula for G(x).
G(1) = 1∫1 (t^2-2)dt = 0 G'(x) = x^2 - 2 G'(1) = (1)^2 - 2 = -1 G'(2) = (2)^2 - 2 = 2 G(x) = x∫1 (t^2-2)dt = (1/3*t^3 - 2t) x|1 = (1/3*x^3 - 2x) - (1/3 - 2) = (1/3)x^3 - 2x - 5/3
4.9. Initial Condition EX. 2: A 900-kg rocket is released from a space station. As it burns fuel, the rocket's mass decreases and its velocity increases. Let υ(m) be the velocity (in meters per second) as a function of mass m. Find the velocity when m = 729 if dυ/dm = −50m^(−1/2). Assume that υ (900) = 0.
Given: dv/dm = a(m) = -50m^(-1/2) v(900) = 0 1. v(m) = ∫a(m)dm = -100m^(1/2) + C 2. -100 * (900)^(1/2) + C = 0 -100 * 30 + C = 0 -3000 + C = 0 C = 3000 v(m) = -100m^(1/2) + 3000 v(729) = -100 * (729)^1/2 + 3000 = -100 * 27 + 3000 = -2700 + 3000 = 3000 m/s
5.1. EX. 2: Calculate the limit using a summation approximation for f (x) = 3x^2 + 4x over [0, 2].
Given: f(x) = 3x^2 + 4x, [0,2] 1. 2/N * (N) Σ (j=1) {(12/N^2)j^2 + (8/N)j} a. Δx = (2 - 0)/N = 2/N b. f(0 + j * 2/N) = f(2/N * j) = 3(2/N * j)^2 + 4(2/N * j) = 3(4/N^2 * j^2) + (8/N * j) = (12/N^2 * j) + (8/N * j) 2. 2/N * (N) Σ (j=1) {(12/N^2)j^2 + (8/N)j} = [2/N * (N) Σ (j=1) (12/N^2)j^2] + [2/N * (N) Σ (j=1) (8/N)j] = [24/N^3 * (N) Σ (j=1) j^2] + [16/N^2 * (N) Σ (j=1) j] = 24/N^3 * (N^3/3 + N^2/2 + N/6) + 16/N^2 * (N^2/2 + N/2) = 24N^3/3N^3 + 24N^2/2N^3 + 24N/6N^3 + 16N^2/2N^2 + 16N/2N^2 = 8 + 12/N + 4/N^2 + 8 + 8/N = 16 + 20/N + 4/N^2 3. lim as N -> inf of (16 + 20/N + 4/N^2) = 16 + 0 + 0 = 16
5.1. EX. 1: Calculate the approximation for the given function and interval. M(4), f (x) = 4x+2, [1, 3]
Given: f(x) = 4x+2, [1,3], N = 4 M(N) = Δx * (N) Σ (j=1) f(a + (j - 1/2)Δx) 1. Δx = (3-1)/4 = 2/4 = 1/2 2. M(5) = 1/2 * (4) Σ (j=1) f(1 + (j - 1/2)(1/2)) = 1/2 * (4) Σ (j=1) f(1 + 1/2j - 1/4) = 1/2 * (4) Σ (j=1) f(3/4 + 1/2j) = 1/2 * [((4(3/4 + 1/2) + 2) + (4(3/4 + 1) + 2) + (4(3/4 + 3/2) + 2) + (4(3/4 + 2) + 2))] = 1/2 * [(3+2+2) + (3+4+2) + (3+6+2) + (3+8+2)] = 1/2 * [7 + 9 + 11 + 13] = 1/2 * 40 = 20
4.9. Initial Condition EX. 1: A particle located at the origin at t = 1s moves along the x-axis with velocity υ(t) = (6t^2 − t) m/s. State the differential equation with initial condition satisfied by the position s(t) of the particle, and find s(t).
Given: v(t) = 6t^2-t s(1) = 0 1. s(t) = ∫v(t)dt = ∫(6t^2-t)dt = 2t^3-1/2t^2+C 2. 2(1)^3-1/2(1)^2+C = 0 2 - 1/2 + C = 0 3/2 + C = 0 C = -3/2 s(t) = 2t^3 - 1/2t^2 - 3/2
5.6. Substitution Method
If F'(x) = f(x), then by the Chain Rule "in reverse": ∫f(u(x))u'(x)dx = F(u(x)) + C = ∫f(u)du Where: f(u) = f(u(x)) du = u'(x)dx 1. Substitute u for the "inner" function of the integrand. u = u(x) 2. Take the derivative with respect to x of this substitution. du/dx = u'(x) 3. Treat du & dx as a "variable" and isolate. du = u'(x)dx **In some cases, the entire derivative u'(x) won't appear exactly as found in the integral, so some manipulation of du or u is required. 4. Plug in the new substitutions and solve. ∫f(u(x))u'(x)dx = ∫f(u)du
5.2. Definition of the Definite Integral
If f (x) is continuous on [a, b], then f (x) is integrable over [a, b]. f(x) is integrable over [a, b] if the limit exists. Integral from [a,b] = Infinite limit of summation b∫a f(x)dx = lim as N -> inf of Δx * (N) Σ (j=1) f(a + jΔx)
5.2. Comparison Theorem
If f(x) >= g(x) for x in [a,b], then: b∫a f(x)dx > b∫a g(x)dx. If m ≤ f (x) ≤ M for x in [a, b], then: m(b-a) ≤ b∫a f(x)dx ≤ M(b-a) M and M are the lower and upper bounds for f(x), whose area lies between two rectangles. Special Case: If f(x) >= 0 for x in [a,b], then b∫a f(x)dx >= 0. Consider: 1. If f(x) - g(x) > 0, then... 2. b∫a [f(x)-g(x) dx > 0 equals b∫a f(x)dx - b∫a g(x)dx > 0
5.1. Area as a Limit of Approximations
If f(x) is continuous on [a,b], then the endpoint and midpoint approximations reach the same limit as N -> infinity. If f(x) >= 0, this is the area for [a,b]. lim as N -> inf of R(N) = lim as N -> inf of L(N) = lim as N -> inf of M(N) lim as N -> inf of Δx * (N) Σ (j=1) f(a + jΔx) To find the area as a limit of an approximation: 1. Find the summation equation. a. Δx = (b - a)/N b. f(a + jΔx) : Plug in and simplify 2. Use summation laws and power sums to find the equation in terms of N. 3. Calculate the limit as N -> inf of the equation.
5.1. Left endpoint rectangles
L(N) = Δx (f (a) + f (a + Δx) + f (a + 2Δx) +...+ f (a + (N − 1)Δx)) L(N) = Δx * (N) Σ (j=0) f(a + (j-1)Δx)
5.5. Integral of Velocity
Let s (t) be the position at time t of an object in linear motion: 1. Then the object's velocity is v (t) = s'(t) 2. The integral of v(t) (s(t)'s rate of change) is equal to the net change in position or displacement over a time interval [t1, t2]. t2∫t1 v(t)dt = s(t2) - s(t1)
5.5. EX. 1: Water flows into an empty reservoir at a rate of 3000 + 20t liters per hour. What is the quantity of water in the reservoir after 5 hours?
Let: r(t) = flow rate = 3000 + 20t q(t) = total quantity of water in reservoir at t hours q(t) = 5∫0 r(t)dt = 5∫0 (3000 + 20t)dt = (3000t + 10t^2) 5|0 = 3000(5) + 10(25) = 15250 liters
5.5. Total vs Marginal Cost
Let: C(x) = Cost of producing x units of a commodity C'(x) = Marginal cost for one additional unit Then: Cost of increasing production from a to b = b∫a C'(x)dx = C(b) - C(a) Total Cost = Marginal Cost + Initial Costs C(0)
5.1. Midpoint rectangles
M(N) = Δx[f (a + 1/2Δx) + f (a + 3/2Δx) +...+ f (a + (N-1/2)Δx)] M(N) = Δx * (N) Σ (j=1) f(a + (j - 1/2)Δx)
5.1. Right endpoint rectangles
R(N) = Δx[f (a + Δx) + f (a + 2Δx) +...+ f (a + NΔx)] R(N) = Δx * (N) Σ (j=1) f(a + jΔx)
5.5. Net Change as the Integral of a Rate
The net change in some function f(t) over an interval [t1, t2] is equal to the area beneath the curve: t2∫t1 f'(t)dt = f(t2) - f(t1) Integral of the rate of change = Net change over [t1, t2]
5.1. Monotonic Functions
When f(x) is monotonic (increasing or decreasing for [a, b]), the exact area lies between R(N) and L(N). f (x) increasing: L(N) ≤ area under graph ≤ R(N) f (x) decreasing: R(N) ≤ area under graph ≤ L(N)
5.4. FTC II and the Chain Rule
Where the upper bound is a continuous function: G(x) = A(g(x)) = g(x)∫a f(t)dt G'(x) = A'(g(x)) * g'(x) = f(g(x)) * g'(x)
5.6. Change of Variables (boundaries)
b∫a f(u(x))u'(x)dx = u(b)∫u(a) f(u)du When using u-substitution, the terms and boundaries of an integral must be accounted for in one of two ways: 1. Use u-substitution to find the antiderivative, then at the end rewrite in terms of x using the original substitution. EG: 3∫2 2xcos(x^2)dx u = x^2 du = 2xdx = 3∫2 cos(u)du = sin(u) 3|2 = sin(x^2) 3|2 <--- Evaluated at this step. 2. Use u-substitution to find the antiderivative, then plug in the old boundaries [a,b] into it to get new boundaries in terms of u, [u(a), u(b)]. The integral can then be solved without rewriting in terms of x. This is usually clearer and faster than method 1. EG: 3∫2 2xcos(x^2)dx u = x^2 u(a) = 2^2 = 4 u(b) = 3^2 = 6 du = 2xdx = 6∫4 cos(u)du = sin(u) 6|4 <--- Evaluated at this step.
5.2. Signed Area
b∫a f(x)dx = Area above - Area below If f(x) >= 0 for [a,b], then b∫a f(x)dx is the area above the x-axis and below the graph y=f(x). If f(x) < 0 for [a,b], then b∫a f(x)dx is the negative area below y=f(x) and the x-axis.
5.3. Absolute Value Integral as Sum of Integrals
b∫a |f(x)|dx 1. Set f(x) <= 0 and solve for x to find the new bound, c, where the function crosses the x-axis (at x = 0). 2. b∫a |f(x)|dx = area above - area below b∫a |f(x)|dx = b∫c f(x)dx - c∫a f(x)dx
5.2. EX. 2: Use the Comparison theorem to show that + 1∫0 x^5dx <= 1∫0 x^4dx but 2∫1 x^5dx >= 2∫1 x^4dx.
f(x) = x^5 g(x) = x^4 f(1/2) = 1/32 g(1/2) = 1/16 f(1/2) < g(1/2) f(x) < g(x) for x in [0,1], thus 0∫1 f(x)dx <= 0∫1 g(x)dx f(3/2) = 243/32 ~= 7.59 g(3/2) = 81/16 ~= 5.06 f(3/2) > g(3/2) f(x) < g(x) for x in [1,2], thus 1∫2 f(x)dx >= 1∫2 g(x)dx
5.7. Integral of Natural Logarithms
x∫1 dt/t = lnx
5.3. EX. 2: Write the integral as a sum of integrals without absolute values and evaluate. π∫0 |cosx|dx
π∫0 |cosx|dx 1. cosx <= 0 = x <= arccos(0) = x <= π/2 2. π∫0 |cosx|dx = π∫π/2 cosxdx - π/2∫0 cosxdx = -sinx π|π/2 - (-sinx) π/2|0 = -sinx π|π/2 + sinx π/2|0 = (0 - (-1)) + (1 - 0) = 1 + 1 = 2
5.7. EX. 3: Evaluate. ∫(3x+2)dx/(x^2+4)
∫(3x+2)dx/(x^2+4) = ∫3xdx/(x^2+4) + ∫2dx/(x^2+4) = (3/2)ln(x^2+4) + arctan(x/2) + C 1. ∫3xdx/(x^2+4) u = x^2+4 du = 2xdx (1/2)du = xdx = (1/2)∫3du/u = (3/2)∫du/u = (3/2)lnu = (3/2)ln(x^2+4) 2. ∫2dx/(x^2+4) = 2∫dx/(x^2+4) = 2∫dx/4(x^2/4+1) = 2∫dx/4((x/2)^2+1) u = x/2 du = 1/2dx 2du = dx = 2∫2du/4(u^2+1) = 4∫du/4(u^2+1) = ∫du/(u^2+1) = arctan(u) = arctan(x/2)
5.7. Integrals of Inverse Trigonometric Functions
∫dx/√(1-x) = arcsinx +C ∫dx/√(1-x²) = -arccosx +C ∫dx/(1+x²) = arctanx +C ∫dx/(1+x²) = -arccotx +C ∫dx/(|x|√(x²-1)) = arcsecx +C ∫dx/(|x|√(x²-1)) = arccscx +C
5.7. EX. 2: Evaluate. ∫e^x(e^(2x)+1)^3dx
∫e^x(e^(2x)+1)^3dx = ∫e^x((e^x)^2+1)^3dx u = e^x du = e^xdx = ∫(u^2+1)^3dx = ∫(u^6 + 3u^4 + 3u^2 + 1)dx = (1/7)u^7 + (3/5)u^5 + u^3 + u = (1/7)e^7x + (3/5)e^5x + e^3x + e^x + C
5.7. Integrals of Exponential Functions
∫e^xdx = e^x + C ∫b^xdx = b^x/lnb + C How did we get this? Look at the derivative of b^x: d/dx b^x = d/dx e^(lnb * x) d/dx b^x = lnb * e^(lnb * x) d/dx b^x = lnb * b^x d/dx (b^x / lnb) = b^x ∫[d/dx (b^x / lnb)] = ∫b^xdx b^x / lnb = ∫b^xdx
5.6. EX. 1: Evaluate the integral. ∫xsin(7x^2+5)dx
∫xsin(7x^2+5)dx u = 7x^2+5 du/dx = 14x du = 14xdx (1/14)du = xdx ∫xsin(7x^2+5)dx = ∫sin(u)*(1/14)du = (1/14)∫sin(u)du = (1/14) * -cos(u) = -(1/14)cos(7x^2+5) + C
