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Aim: How do we find anti-derivatives of elementary functions? 2-day lesson Objectives: to find the anti-derivatives of trig, exponential and inverse trig functions, to find the particular solution, to find the anti-derivative of a product or quotient of polynomials Grouping: students are given opportunity to work in cooperative setting during this class. Time is purposely set aside for students to work collaboratively on at least one set of exercise problems. During this time, students are encouraged to explain concepts and materials and solutions to each other. Furthermore, students get to present their solutions on the board. Differentiated instruction: all students are held to the highest standard. However the students that are not performing well are grouped with students that are excelling to serve as study partners. Teacher will informally assess their understanding by asking them questions to ensure that they are keeping up. Assessment: Class is given multiple occasions to work individually and in groups. Teacher circulates the room to assist and also assess student understanding. Furthermore, students are encouraged to explain work to each other. This is another opportunity for teacher to assess student understanding. Lastly, when lesson is finished, students are grouped for class work, questions that are designed as exit slip problems. Homework Review (10 minutes): Students are assigned problems to present on board. Students will also answer questions. Teacher can go over a problem if no one in class could explain it Lesson Development: [15 – 20 min] Just as there are several ways to mean differentiation—differentiate, taking the derivative, or finding the instantaneous rate of change, there are several ways to mean integration—indefinite integral, antiderivative, or solve the differential equation* Example: What does solving the differential equations mean? We know that asks us to find the function when given its derivative function. “Solve the differential equation” means the same except sometimes it comes with an initial condition which enables us to find the value of c. Example: a) f '( x) 4 x, f (0) 6 b) h '(t ) 8t 3 5, h(1) = -4 x2 f ( x) 4 xdx 4[ ] c 2 x 2 c 2 2 f (0) 2(0) c 6 c6 t4 h(t ) (8t 5)dt 8( ) 5t c 4 4 h(1) 2(1) 5(1) c 4 c 11 Particular Solution: f ( x) 2 x 2 6 Particular Solution: h(t ) 2t 4 5t 11 3 c) f ''( x) x 3/2 , f '(4) 2, f (0) 0 f '( x) x 3/2 dx x 1/2 c (1/ 2) f '(4) 2(4) 1/2 c 2, c = 3 f '( x) 2( x) 1/2 3 x1/2 f ( x) (2 x 3)dx 2( ) 3x c 1/ 2 f (0) 4(0)1/2 3(0) c 0, c = 0 1/2 Particular Solution: f ( x) 4 x1/2 3 x d) f ''( x) sin x, f '(0) 1, f (0) 6 f '( x) sin xdx cos x c f '(0) cos 0 c 1, c 2 f '( x) cos x 2 f ( x) ( cos x 2)dx sin x 2 x c f (0) sin 0 2(0) c 6, c = 6 Particular Solution: f ( x) sin x 2 x 6 EX1: An evergreen nursery usually sells a certain shrub after 6 years of growth and shaping. The growth dh 1.5t 5 , where t is the time in years and h is the height rate during those 6 years is approximated by dt in centimeters. The seedling are 12 centimeters tall when planted (t=0). [10 – 12 min] a) Find the height after t years t2 h (1.5t 5)dt 1.5 5t c 2 To solve for c, we plug the initial condition (0, 12) back to the equation we just found: 1.5 2 12 (0) 5(0) c 2 c 12 h 1.5 b) how tall are the shrubs when they are sold? The shrubs are sold after 6 years, so t = 6. h 1.5 (6) 2 5(6) 12 1.5(18) 42 69cm 2 t2 5t 12 2 Going forward, we will use the dominant terms--indefinite integral, anti-derivative--interchangeably. [8 – 10 min] We know that: d d d sin x cos x, cos x sin x, tan x sec 2 x, dx dx dx d x d 1 d d e ex , ln x , a x ln a( a x ), sec x sec x tan x dx dx x dx dx d 1 d 1 tan 1 x 2 , sin 1 x dx x 1 dx 1 x2 sin xdx cos x c, cos xdx sin x c, sec Therefore: x x e dx e c, x a ax , ln a 2 1 x dx ln x c, x sec x tan xdx sec x c, 1 1 x2 EX1: If f '( x) e x 3 and f (0) 2 , what is f ( x) ? [5 min] xdx tan x c 1 dx tan 1 x c 1 2 dx arcsin x c EX2: If f '( x) 3 x and f (1) 2 , what is x f ( x) ? [5 min] 3 x2 xdx 3ln x c x 2 1 f (1) 3ln1 c 2 2 3 c 2 x2 3 f ( x) 3ln x 2 2 f ( x) e x 3dx e x 3x c f ( x) f (0) e0 c 2 c 1 f ( x) e x 3x 1 Is ( x 1)( x 2)dx ( x 1)dx ( x 2)dx ? Verify it using wolframalpha.com How can we integrate this? Since we know f ( x) g ( x)dx f ( x)dx g ( x)dx , why don’t we turn this product into a sum or difference? How? 2 ( x 1)( x 2)dx ( x x 2)dx x3 x 2 2x c 3 2 Same is true if we are asked to integrate a quotient. x3 2 x 1 1 x3 2 x dx x 2 x dx 3 2 x ln x c EX3: Integrate [10 – 12 min] 1 y2 a) dy y b) ( x 2) 2 dx 1 y 1 y dy ydy ln y c y y 2 2 2 2 2 ( x 2) dx x 4 x 4dx x3 2x2 4x c 3 c) 2 x5 8 3/2 x ( x 2)dx x3/2 2 x1/2 dx x c 5 3 x ( x 2)dx d) ( x 2) dx x x1/2 2 x 1/2 dx 2 x3/2 x1/2 c 3 HW#44: P291 – 293: 23 – 28, 31, 32, 34 – 37, 39, 41 – 44, 63, 66, 69, 70, 74 (due in two classes since this is a 2-day lesson) HW#44 Solutions: P291 – 293: 23 – 28, 31, 32, 34 – 37, 39, 41 – 44, 63, 66, 69, 70, 74 23) 24) 25) 26) 27) 28) 31) 32) 34) 36) 39) 35) 37) 41) 42) 44) 43) 63) 69) 66) 70) 74)