Vector integrals Line integrals Surface integrals Volume integrals Integral theorems
Ch. 5 Review: Integrals
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Transcript of Ch. 5 Review: Integrals
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Ch. 5 Review: Integrals
AP Calculus
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5.2: The Differential dy5.2: Linear Approximation5.3: Indefinite Integrals5.4: Riemann Sums (Definite Integrals)5.5: Mean Value Theorem/Rolle’s Theorem
Ch. 5 Test Topics
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dx & dy: change in x and y for tangent (derivative)
The Differential dy
Tangent line
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Find the differential dy:y =
dy = (6x – 4) dx
𝑑𝑦𝑑𝑥
= 𝑓 ′ (𝑥 ) , 𝑠𝑜𝑑𝑦= 𝑓 ′ (𝑥) ∙𝑑𝑥
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Linear Approximation
Write the equation of the line that bestfits at x = 2. Then find dx, and dy if f(2.01) is approximated.
Equation:
dx
dy
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Linear Approximation
Write the equation of the line that bestfits 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 –
y + 2 = 5(x – 2) y = 5x – 12 If x = 2.01, y = -1.95
: Function change in y: f(2.01) – f(2) = .0503dx: Tangent line change in x -- 2.01 – 2 = .01dy: 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
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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
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If a function f is:1) Differentiable for all values of x in the
open interval (a, b) and2) 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
Mean Value Theorem (MVT)
f’(c) =
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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
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Remember – Function must be CONTINUOUS and DIFFERENTIABLE on interval! Otherwise, conclusion of MVT may not be met.
Mean Value Theorem
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Integrals Self-Quiz
∫ 8𝑥1 /3𝑑𝑥=¿¿∫ (5𝑥4+1 )𝑑𝑥=¿¿∫(7 𝑥+3)8𝑑𝑥=¿¿
∫5 𝑠𝑖𝑛2 𝑥 𝑑𝑥=¿¿
∫ 𝑠𝑒𝑐 5 𝑥 tan5 𝑥 𝑑𝑥=¿¿
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Integrals Self-Quiz
∫ 8𝑥1 /3𝑑𝑥=6 𝑥4 /3+𝑐∫ (5𝑥4+1 )𝑑𝑥=𝑥5+𝑥+𝑐∫(7 𝑥+3)8𝑑𝑥=
163
(7𝑥+3)9+𝑐
∫5 𝑠𝑖𝑛2 𝑥 𝑑𝑥=−52cos2𝑥+𝑐
∫ 𝑠𝑒𝑐 5 𝑥 tan5 𝑥 𝑑𝑥=15sec 5 𝑥+𝑐
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Integrals Self-Quiz
∫𝑒sin 𝑥𝑐𝑜𝑠𝑥 𝑑𝑥=¿¿
∫𝑥 (𝑥2−3)5 𝑑𝑥=¿¿
∫𝑐𝑜𝑠4 𝑥𝑠𝑖𝑛𝑥 𝑑𝑥=¿ ¿
∫ 2𝑥 (𝑥3−7 )𝑑𝑥=¿¿
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Integrals Self-Quiz
∫𝑒sin 𝑥𝑐𝑜𝑠𝑥 𝑑𝑥=𝑒𝑠𝑖𝑛𝑥+𝑐
∫𝑥 (𝑥2−3)5 𝑑𝑥=112
(𝑥2−3)6+𝑐
∫𝑐𝑜𝑠4 𝑥𝑠𝑖𝑛𝑥 𝑑𝑥=−15𝑐𝑜𝑠5𝑥+𝑐
∫ 2𝑥 (𝑥3−7 )𝑑𝑥=25𝑥5−7 𝑥2+𝑐
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R Problems, pg. 260: R1 –R5 ab