MATH Section 210

Transcription

1 MATH Section 10 Instructor: Avner Segal (avners@math.ubc.ca) January 31 st 017 Common course page: Individual section page: Office hours: Thursdays 1:00-14:30 (LSK300) On Quiz Weeks: Tuesdays 1:00-13:00 (LSK300) - by appointment only! Follow announcements on section web page in case of changes.

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3 Quiz # 1 (V1) - Problem x Define f (x) = cos(t )dt. x (a) Find f (1). Simplify your answer. (b) Find f (1). Simplify your answer.

4 x Define f (x) = cos(t )dt. x.(a) Find f (1). Simplify your answer. By definition f (1) = a a g(x)dx = cos(t )dt = 0, because always.(b) Find f (1). Simplify your answer. Write F (x) = x 0 cos(t )dt, It holds that F (x) = cos(x ). So, x x x f (x) = cos(t )dt = cos(t )dt cos(t )dt = F (x) F (x ). x 0 0 It follows that f (x) = F (x) (x ) F (x ) = cos(x ) x cos(x 4 ). In particular, f (1) = cos(1) cos(1) = cos(1). Many have missed the inner derivative in this exercise.

5 Quiz # 1 (V1) - Problem 3 (a) For a certain function f (x) and a certain number b, the following equation holds: lim Find f (x) and b. n i=1 ( n 3 n i 1 ) 3 n 1 b = f (x)dx (b) Evaluate b f (x)dx, using the f (x) and b you found in part (a). Simplify your answer.

6 lim n i=1 3.(a) Find f (x) and b ( n 3 n i 1 ) 3 n 1 b = f (x)dx We first note that the term ( i ) 1 3 n is a hint that we should consider the midpoint rule. We also take x = 3 n. We then recall that x = b a n and that a =, it follows that b = 5. We now look for the function f (x). Using the midpoint rule (( f i 1 ) ) ( 3 n + = f (x i,n ) = i 1 ) 3 n 1 (i = 1 ) n. We conclude that f (x) = x 3. Many of you forgot to write the functions in terms of the midpoint rule (or any other for that matter) and got x 1 instead of x 3.

7 5 3.(b) Evaluate x 3 dx The integral is the sum of areas of the two triangles. Namely, 5 x 3 dx = = 5. Alternatively, one can write 5 x 3 dx = 3 (3 x)dx (x 3)dx =... = 5. Some, accidentally, dropped the absolute value and solved for 5 (x 3)dx.

8 Quiz # on Thursday (February nd ) Material covers everything up to (and including) integration by parts. Don t forget your UBC Card and a black pen!

9 Trigonometric Substitution We aim to simplify (and solve) integrals containing expressions of the form a x x +a x a a x x x a θ θ θ a 1 sin θ = cos θ tan θ +1 = sec θ sec θ 1 = tan θ

10 Trigonometric Substitutions 1. In order to simplify a x we use the substitution x = asinu (dx = cosudu). We then have a x = a a sin u = a cos u = acosu = a cosu.. In order to simplify x +a we use the substitution x = atanu (dx = sec udu). a +x = a +a tan u = a sec u = asecu = a secu. 3. In order to simplify x a we use the substitution x = asecu (dx = secu tanudu). x a = a sec u a = a tan u = atanu.

11 Question x dx x 1 =? Solution x { } dx x 1 = x = secu sec u = secu tanudu = dx = secu tanudu sec u 1 sec 3 u tanu tanu du. Assume now that x = secu > 1, then tanu > 0 and we get x dx x 1 = sec 3 udu (fact) = 1 secu tanu + 1 log secu +tanu +C Applying tan(arcsecu) = x 1 we get = 1 x x 1+ 1 x log + x 1 +C. What about x < 1? (Note that x 1 is not defined for 1 < x < 1)

12 Differentiating this shows that this is an antiderivative of x x 1 for x < 1 two. ( d 1 dx x x 1+ 1 x log ) + x 1 x 1 x = + x ( ) 1 x + x 1 + x 1 x 1 x 1 x = + x 1 + x + x 1 (x + x 1) x 1 = x 1 x 1 + x x x 1 = x x 1.

13 Question 5 x x 3 =? x 1 3 Solution 5 x x 3 5 (x 1) 4 = dx 3 x 1 3 x 1 We can first make a substitution u = x 1 and then do the trigonometric substitution, but we will now do it in one go. Write x = 1 +secu. So dx = secu tanudu. But what about the limits of integration? Should solve the equation 1 +secu = 3 and 1 +secu = 5. The solutions are 0 and π/3 respectively. 5 3 π/3 tan udu = 0 x x 3 π/3 (secu) 4 = secu tanudu x 1 0 secu π/3 0 ( sec u 1 ) du = [tanu u] π/3 ( π ) 0 = 3. 3

14 4 1. (Final, 014) Evaluate x dx. 1 dx. (Final, 013) Evaluate 1 (x +1) (MATH final, a variant of MATH final) Evaluate (3 x x ) 3/ dx. 4. (Final, 008) Find the area inside the ellipse x a + y b = Evaluate r a x r x dx

15 1. Solution - 4 x dx Let x = sinθ so that dx = cosθdθ. Then 4 x dx = 4 4sin θcosθdθ cos = 4 θ cosθdθ = 4 cos θdθ 1 +cos(θ) 4 dθ = θ +sin(θ) +C = θ +sin(θ) +C = θ +sin(θ)cos(θ) +C ( x ) = arcsin +x 1 x 4 +C. We used θ = arcsin ( ) x, sinθ = x, cosθ = ± 1 sin θ.

16 1 dx. Solution - 1 (x +1) 3 Let x = tanθ so that dx = sec θdθ. Then (using cos ϕ = 1+cos(ϕ) ) 1 1 dx π/4 (x +1) 3 = sec θdθ π/4 π/4 (tan θ +1) 3 = sec θdθ π/4 sec 6 θ π/4 π/4 ( ) 1 +cos(θ) = cos 4 θdθ = dθ π/4 π/4 π/4 = π/4 1 +cos(θ) +cos (θ) dθ 4 π/4 [ 1 = π/4 4 + cos(θ) [ 3θ = 8 + sin(θ) cos(4θ) ] dθ 8 + sin(4θ) ] π/4 = 3π 3 θ= π/

17 3. Solution - (3 x x ) 3/ dx Note that 3 x x = 4 (x +1). Let x = sinθ 1 so that dx = cosθdθ. Then (3 x x ) 3/ dx = (4 (x +1) ) 3/ dx = = (4 4sin θ ) 3/ cosθdθ = (4cos θ ) 3/ cosθdθ 1 8 cos 3 θ cosθdθ = 1 4 dθ cos θ = tanθ 4 +C = 1 4 x +1 +C. 4 (x +1) We used the fact that π θ π and hence cosθ 0. Also, since sinθ = x+1 x+1 we have tanθ =. 4 (x+1), θ 4 (x +1) x +1

18 4. Solution - Area of ellipse x a + y b = 1 Assume that a,b > 0. We look for the area between the curves y = b 1 x and y = b 1 x for a x a. a a Let x = asinθ so that dx = acosθdθ and π θ π. Then, the area is given by a A = b 1 x a π = ab π a dx = b π π cos θ cosθdθ = ab π 1 sin θacosθdθ π cos θdθ =... = πab Note that when a = b this is a circle and the formula fits the area of a circle of radius a = b.

19 5. Solution - r a x r x dx After all of this example, we opt to try x = sinu. This would lead us to the integral arcsin(r) arcsin(a) &#@?% sinu cos udu = {z = cosu} = z dz =... $&!#$ I suppose it is tempting, if the only tool you have is a hammer, to treat everything as if it were a nail (Abraham Maslow, 1966). Warning: Trying to hammer a thumbtack can result in injury. r x r x dx = { u(x) = r x } = 1 a 1 [ ] u 3/ 0 = 1 ( r a ) 3/. 3/ 3 r a 0 r a udu =

20 Partial Fractions dx dx = log x +C, = arctanx +C, x 1 +x dx x +3 = 1 dx log x +3 +C, 3 +4x = 1 ( ) x 3 arctan +C, 3 x 3 ( +x x 1 dx = x + 1 x )dx = x +log x +1 +log x 1 +C. x 1 xdx x +x + = 1 (x +)dx x +x + dx (x +1) +1 = 1 log x +x + arctan(x +1) +C. We wish to use similar methods in order to solve integrals of general rational functions (i.e. a quotient of two polynomials).

21 The Idea - Partial Fractions! Given a rational function N(x) (where the numerator N(x) and the D(x) denominator D(x) 0 are polynomials) we will write as a sum of functions of the following kinds Polynomials (when the degree of N(x) is bigger than the degree of D(x)). A, where A (ax+b) n 0, a 0 and b are constants and n is a positive integer. Ax+B, where A, B (not both zero), a 0, b and c are (ax +bx+c) n constants such that ax +bx +c 0 for any x and n is a positive integer. How to do it? We will se in a series of examples. What is it good for? We already know how to integrate all of this functions. So we proceed in two steps: 1. Decomposition into partial fractions.. Integration of the partial fractions.

22 Remark - Irreducible Polynomials A polynomial p(x) is called irreducible if it cannot be written as a product of two non-constant polynomials, p(x) = p 1 (x)p (x). Fact: Irreducible polynomials can take only one of this forms: Linear, ax +b. Quadratic ax +bx +c with negative discriminant, i.e. = b 4ac < 0. This are exactly the quadratic polynomials satisfying p(x) 0 for all x.

23 Question Which one of the following is irreducible? A. x +7. B. x 7. C. x 4. D. x +3x 4. E. x +3x +4.

24 Question Which one of the following is irreducible? A. x +7, = 8 < 0. B. x 7, = 8 > 0. C. x 4, (proper) linear polynomials are always irreducible. D. x +3x 4, = 9 +3 = 41 > 0. E. x +3x +4, = 9 3 = 3 < 0.