6.1 The Indefinite Integral

1. 6.1 The Indefinite Integral • An Antiderivative of a function f is a function F such that F = f. • X4 is an antiderivative of 4x3 • X4 + 45 is also an antiderivative of 4x3 • Indefinite integral of a function : set of allantiderivatives of that function Because the derivative of x4 is 4x3 • And the derivative of x4 + 45 is 4x3 • f (x)dx : “the indefinite integral of f (x) with respect to x” • the set of all antiderivatives of f. • f is called the integrand, and x is called the variable of integration. 4x3dx = x4 + C Everypossibleantiderivative of 4x3 has the form x4 + C. • Check that

2. Rules for Indefinite Integrals • Power Rule for the Indefinite Integral • Rules for Exponential Functions • Rules for Sums/Differences --- Constant Multiples --- Absolute Value • Examples • (if n ≠ –1)  [f(x)  g(x)]dx = f(x)dx  g(x)dx  |x| dx = x |x| + C 2  k f(x) dx = k f(x)dx

3. Application: Finding Cost from Marginal Cost • The marginal cost to produce baseball caps at a production level of x caps is 4 – 0.001x dollars per cap, and the cost of producing 100 caps is $500. Find the cost function. • Recall : Marginal cost function is the derivative of the cost function • So, C(x) = 4 – 0.001xMust find C(x). • Now we must find K • C(x) = (4 – 0.001x)dx • C(100) = 500 • C(100) = 4(100) – 0.0005(100)2 + K • 500 = 395 + K • K = 105. • C(x) = 4x – 0.0005x2 + 105

4. Motion in a Straight Line • Position, Velocity, and Acceleration: Integral Form • Example • Velocity of a particle moving in a straight line is given by v(t) = 4t + 1 • => position after t seconds is s(t) = v(t)dt = (4t + 1)dt = 2t2 + t + C.

5. Another Example • The velocity of a particle moving along in a straight line is given by • v(t) = 4t + 1 m/s. The particle is at position s = 2 meters at time t = 1. Find an expression for s in terms of t. • s(t) = v(t)dt = (4t + 1)dt S(1) = 2 • S(t) = 2t2 + t + C 2 = 2(1)2 + 1 + C • C = -1 • Thus,s(t) = 2t2 + t – 1 meters • For a freely falling body experiencing no air resistance and zero initial velocity, find an expression for the velocity v in terms of t. [Note: On Earth, a freely falling body experiencing no air resistance accelerates downward at approximately 9.8 m/s2 (or 32 ft/s2).] • a(t) = –9.8 m/s2 • V(0) = 0 • 0 = –9.8(0) + C • C = 0 • v(t) = a(t)dt • =(–9.8)dt • =–9.8t + C. • v(t) = –9.8t m/s

6. Motion in a Straight Line • Vertical Motion Under Gravity: Velocity and Position • If one ignores air resistance, the vertical velocity and position of an object moving under gravity are given by • British UnitsMetric Units • Velocity: v(t) = –32t + v0 ft/s v(t) = –9.8t + v0 m/s • Position: s(t) = –16t 2 + v0t + s0 ft s(t) = –4.9t 2 + v0t + s0 m • v0 = initial velocity = velocity at time 0 • s0 = initial position = position at time 0 • Example • If a ball is thrown down at 2 ft/s from a height of 200 ft, then its velocity and position after t seconds are • v(t) = –32t – 2 ft/s and s(t) = –16t 2 – 2t + 200 ft.

7. 6.2 Substitution Technique for Integration • The chain rule for derivatives gives us the antiderivative technique. • Recall : to differentiate a function like (x2 + 1)6, we first think of the function as g(u) where u = x2 + 1 and g(u) = u6. We then compute the derivative, using the chain rule: • Thus, d/dx (x2 + 1)6 = 6(x2 + 1)5 (2x) = 12x (x2 + 1)5 Suppose we want to evaluate an integral such as: ∫ 4x(x2 + 1)6dx. Substitution Method 1. Write u as a function of x. 2. Take the derivative du/dx and solve for the quantity dx in terms of du. 3. Use the expression you obtain in step 2 to substitute for dx in the given integral and substitute u for its defining expression.

8. Example: – Substitution • Find ∫ 4x(x2 + 1)6dx. • Step 1: Choose an expression for “u”. Often an expression raised to a power is the best choice. Write u as a function of x. • Step 2:Take the derivative of u with respect to x • Step 3:Solve for dx Step 4: Substitute u for the defining expression and dx in the original integral: = ∫2u6du Step 5: Solve the integral and Substitute back in for u for the final answer ∫4x(x2 + 1)6dx = ∫4xu6du

9. Shortcuts for (ax + b) expressions • Rule Quick Example

10. When to Use Substitution & What to use for “u” See FAQs on P. 476 in your Text for an Excellent Resource In General: Pretend you want to take the derivative instead of the integral. If taking the derivative would require chain rule then integrating May require substitution. There are no rules for choosing “u” Common choices for “u” are: • A linear expression raised to a power: Set u = the linear expression (3x -2)-3 => Let u = 3x – 2 • A constant raised to a linear expression: Set u = the linear expression 32x+1 => Let u = 2x + 1 There are several more on page 476. Check them out!

11. 6.3 The Definite integral The Indefinite Integral, has many solutions with infinitely many “Constants” of Integration and the integral is not defined for any particular interval. The Definite Integral IS defined for a particular interval [a, b] and has only one solution - a number. Graphically: The definite integral gives the area under a curve from [a, b] We form rectangles under the curve, dividing the area into a number of subdivisions, then add the rectangle areas together to estimate the total area under the curve and thus, the definite integral. Numerically: We use a formula to calculate the sum of the subdivision areas We call this a Riemann Sum. (Note: There are “left” and “right” riemann sums) Check out this website for some clear pictures showing the difference:http://www2.seminolestate.edu/lvosbury/CalculusI_Folder/RiemannSumDemo.htm

12. Example1: Total Cost • A cell phone company’s pricing scheme marginal cost of a phone call is • dollars per hour. Estimate the total cost of • a 2 hour phone call. (Left Riemann Sum Examples) • Note: x = (b – a)/n = f(x0)x + f(x1)x +· · ·+ f(xn–1)x. = (xk)x 5 x 2 = $10 5(1) + .45(1) = $5.45 Minute-by-Minute (120 subdivisions) Cost ≈ $1.56 Second-by-Second (7,200 subdivisions) Cost ≈ $1.52 5(.5) + .83(.5) +.45(.5) + .31(.5)= $3.30 5(.25) + 1.43(.25) + .83(.25) + .59(.25) +… = $2.31

13. The Definite Integral: Numerical and Graphical Approaches • Interpretation of the Riemann Sum • If f is the rate of change of a quantity F (that is, f = F ), then the Riemann sum of f approximates the total change of F from x = a to x = b. (The approximation improves as the number of subdivisions increases toward infinity.) • Interpretation Examples • If f (t) is the rate of change in the number of bats in a belfry and [a, b] = [2, 3], then the Riemann sum approximates the total change in the number of bats in the belfry from time t = 2 to time t = 3. • If c(x) is the marginal cost of producing the xth item and [a, b] = [10, 20], then the • Riemann sum approximates the cost of producing items 11 through 20.

14. The Definite Integral: Numerical and Graphical Approaches • The Definite Integral • If f is a continuous function, the definite integral of f from a to b is defined to be the limit of the Riemann sums as the number of subdivisions approaches infinity: If f is the rate of change of a quantity F (that is, f = F), then is the (exact) total change of F from x = a to x = b.

15. 6.4 The Definite Integral: Algebraic Approach • The Fundamental Theorem of Calculus (FTC) • Let f be a continuous function defined on the interval [a, b] and let F be any antiderivative of f defined on [a, b]. Then • Moreover, an antiderivative of f is guaranteed to exist.

16. Application: Total Cost • Your cell phone company offers you an innovative pricing scheme. When you make a call, the marginal cost is • dollars per hour. • Compute the total cost of a 2-hour phone call.