Lesson 5-5c
This lesson focuses on understanding the U-substitution technique in integral calculus. Students will learn when to apply U-substitution, solve various integrals involving algebraic, exponential, logarithmic, and trigonometric functions, and utilize symmetry properties of functions to simplify integration around the y-axis. Key vocabulary includes even functions, odd functions, and change of variables. Example problems reinforce the concepts covered, allowing students to practice their skills. Homework assignments will strengthen understanding of U-substitution and symmetry in integrals.
Lesson 5-5c
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Presentation Transcript
Lesson 5-5c U-Substitution or The Chain Rule of Integration
Quiz ∫ cos (3x) dx = • Homework Problem: • Reading questions: Which is even, which is odd?
Objectives • Recognize when to try ‘u’ substitution techniques • Solve integrals of algebraic, exponential, logarithmic, and trigonometric functions using ‘u’ substitution technique • Use symmetry to solve integrals about x = 0 (y-axis)
Vocabulary • Change of Variable – substitution of one variable for another in an integral (sort of reverse of the chain rule) • Even Functions – when f(-x) = f(x); even functions are symmetric to the y-axis • Odd Functions – when f(-x) = -f(x); odd functions are symmetric to the origin
Integrals of Symmetric Functions • If a function, f(x), is even [f(-x) = f(x)], then its integral from –a to a is • If a function, f(x), is even [f(-x) = -f(x)], then its integral from –a to a isbecause of signed area (above axis and below) cancel each other out. a a a ∫ ∫ ∫ -a 0 -a f(x) dx = 2 f(x) dx f(x) dx = 0
Example 1 = ∫(2x + 1)² 2/2 dx ∫(2x + 1)² dx If we let u = 2x +1then it becomes u² and du = 2dx we are missing a 2 from dxso we multiple by 1 (2/2) = ½ ∫(2x + 1)² 2 dx ½ goes outside ∫ and 2 stays with dx = ½ ∫u² du = 1/6 u³ + C = 1/6 (2x + 1)³ + C
Example Problems ∫ 1) (2x +3) cos(x² + 3x) dx Find the derivative of each of the following: ∫ • (5x² + 1)² (10x) dx Let u = x² + 3x then du = 2x + 3 So it becomes cos u du Let u = 5x² + 1 then du = 10x So it becomes u² du ∫ ∫ = cos u du = sin (u) + C = u² du = ⅓ u³ + C = sin (x² + 3x) + C = ⅓ (5x² + 1)³ + C
Example Problems cont 4 ∫ 3) tan²(t) cot (t) dt Find the derivative of each of the following: π/4 x ∫ t tan (t)dt x
x² ∫ 6) √2 + sin (t) dt 2 Example Problems cont 2x ∫ 5) 5t sin (t) dt Find the derivative of each of the following: 1
Example Problems cont x³ ∫ 7) √1 + t4dt Find the derivative of each of the following: cos x x ∫ 8) t² dt sin x
Summary & Homework • Summary: • alksdfj • Homework: • Day One: pg 420 - 422: 1, 2, 6, 8, 13, 21, 35, 42, 51, 58, 59, 76 • Day Two: pg 420 - 422: 1, 2, 6, 8, 13, 21, 35, 42, 51, 58, 59, 76
