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Ch. 5 Review: Integrals { AP Calculus Ch. 5 Test Topics 5.2: The Differential dy 5.2: Linear Approximation 5.3: Indefinite Integrals 5.4: Riemann Sums (Definite Integrals) 5.5: Mean Value Theorem/Rolle’s Theorem Tangent line dx & dy: change in x and y for tangent (derivative) ∆ ∆: ℎ The Differential dy = ′ , = ′() ∙ Find the differential dy: y = 3 2 − 4 + 2 dy = (6x – 4) dx Write the equation of the line that best fits = 3 2 − 7 at x = 2. Then find ∆, ∆, dx, and dy if f(2.01) is approximated. ∆ Equation: ∆ dx dy Linear Approximation Write the equation of the line that best fits = 3 2 − 7 at x = 2. Then find ∆, ∆, dx, and dy if f(2.01) is approximated. Point of tangency: f(2) = -2 Slope of tangent (deriv): y’ = 6x – 7 when x = 2 5 Sub into pt-slope equation: y – 1 = ( − 1 ) y + 2 = 5(x – 2) y = 5x – 12 If x = 2.01, y = -1.95 ∆: Function change in x: 2.01 − 2 = .01 ∆ : Function change in y: f(2.01) – f(2) = .0503 dx: Tangent line change in x -- 2.01 – 2 = .01 dy: Tangent line change in y for x = 2 to 2.01: -1.95 - -2 = .05 or dy = f’(x) dx at x = 2 (6(2) – 7)(.01) = .05 Linear Approximation If a function is continuous and differentiable on the interval [a, b], then there is at least one point x = c at which the slope of the tangent equals the slope of the secant connecting f(a) and f(b) Mean Value Theorem If a function f is: 1) Differentiable for all values of x in the open interval (a, b) and 2) Continuous for all values of x in the closed interval [a, b] Then there is at least one number x = c in (a, b) such that f’(c) = −() − Mean Value Theorem (MVT) If a function is differentiable and continuous on the interval [a, b], and f(a) = f(b) = 0, then there is at least one value x = c such that f’(c) = 0. Rolle’s Theorem Remember – Function must be CONTINUOUS and DIFFERENTIABLE on interval! Otherwise, conclusion of MVT may not be met. Mean Value Theorem 8 1/3 = 5 4 + 1 = (7 + 3)8 = 52 = 5 tan 5 = Integrals Self-Quiz 8 1/3 = 6 4/3 + 5 4 + 1 = 5 + + (7 + 3)8 1 = (7 + 3)9 + 63 5 52 = − cos 2 + 2 1 5 tan 5 = sec 5 + 5 Integrals Self-Quiz sin = ( 2 − 3)5 = 4 = 2 3 − 7 = Integrals Self-Quiz sin = + ( 2 − 3)5 1 2 = ( − 3)6 + 12 1 = − 5 + 5 4 2 3 2 5 − 7 = − 7 2 + 5 Integrals Self-Quiz R Problems, pg. 260: R1 –R5 ab