Assignment 13 Assigned Mon Oct 4

Transcription

1 Assignment 3 Assigned Mon Oct 4 We refer to the integral table in the back of the book. Section 7.5, Problem 3. I don t see this one in the table in the back of the book! But it s a very easy substitution x x = = u + du u u + u du = 3 u3/ + 4u / + C = 3 (x )3/ + 4(x ) / + C. x = u +, = du Section 7.5, Problem 9. Use Integral #5 with a = to get x 4x x = (x + )(x 6) 4x x 6 = (x + )(x 3) 4x x ( x sin + 4 sin ( x ) + C ) + C. Section 7.5, Problem 5. Use Integral #8 with a = 3 and b = 3 to get e 3t cos 3t dt = e3x e3x 3 (3 cos 3x + 3 sin 3x) + C = (cos 3x + sin 3x) + C

2 Section 7.5, Problem 9. Use Integral # with n = and a = to get x tan x = x3 3 tan x 3 x 3 + x. So now we have to do a new integral. Dividing x 3 by + x gives so x 3 + x = x x 3 = x x + x, x + x = x ln + x + C. Substituting this in above gives x tan x = x3 3 tan x 6 x + 6 ln + x + C. Section 7.5, Problem 5. Use Integral #6(c) with a = 3 and b = 4 and x = θ to get cos θ 3 cos θ 4 dθ = sin( 3 4 )θ ( 3 4 ) + sin( )θ ( ) + C = 6 sin θ + θ sin C. 4

3 Section 7.5, Problem 9. First make a substitution, sin x = u sin u du x = u, = u du Now use Integral #99 with n = and a = and x = u to get u sin u du = u sin u u du u This last integral can be evaluated using Integral #33 with a = and x = u, u du = u sin u u u + C. Putting it all together gives u sin u du = u sin u 4 sin u + 4 u u + C. Then we substitute back with u = x (and remember there was an extra in the original formula) to get sin x = x sin x sin x + x x + C ( = x ) sin x + x x + C Section 7.5, Problem 3. First make the subsitution x = u to get x u = u du = u du. x u u x = u, = u du Now use Integral #33 with a = and x = u to get u du = u sin u u u + C. Now substituting back u = x (and remembering the extra ) gives x = sin x x x + C = sin x x x + C. x 4

4 Section 7.5, Problem 39. Use Integral #68, which is a reduction formula, with n =, m = 3, a =, to get sin θ cos 3 θ dθ = sin θ cos4 θ + cos 3 θ dθ. 5 5 Next we use Integral #6 with n = 3 and a = to reduce the new integral, cos 3 θ dθ = cos θ sin θ + cos θ dθ 3 3 = cos θ sin θ + sin θ + C. 3 3 Combining this with the earlier formula gives the desired solution, sin θ cos 3 θ dθ = sin θ cos4 θ + cos θ sin θ + sin θ + C Section 7.5, Problem 59. The function f(x) = x x is non-negative for x, and it is negative for x < and for x >. So in the integral b a x x we re not allowed to take a < or b >. Since the integrand is positive, the largest possible value is x x The book doesn t ask you to evaluate it, but you can use Integral #48 with a = to get x x = x x x + ( x ) 8 sin + C = x 4 x x + 8 sin (x ) + C. Then x x = 8 sin () 8 sin ( ) = π. 43

5 Assignment 4 Assigned Weds Oct 6 Section 7.6, Problem 3. Part I Trapezoidal Rule: We divide the interval from to into 4 pieces, so x = 4 =.5, and our x values are x =, x =.5, x =, x 3 =.5, x 4 =. The y values are (note y = f(x) = x + ) y =, y =.5, y =, y 3 =.5, y 4 =. Then the trapezoidal rule gives T = x (y + y + y + y 3 + y 4 ) =.5 ( ) =.75. We have f (x) =, so the maximum value of f on the interval is, so the error estimate satisfies E T The actual value of the integral is M(b a)3 n = ( ( )3 ) 4 = (x + ) = So the actual value of the error is ( ) x= 3 x3 + x = 8 x= 3 = E T = (true value) (trapezoidal value) = = So E T = = 3.5%. (true value)

6 Part II Simpson s Rule: The first part is the same. We divide the interval from to into 4 pieces, so x = 4 =.5, and our x values are x =, x =.5, x =, x 3 =.5, x 4 =. The y values are (note y = f(x) = x + ) Then Simpson s rule gives y =, y =.5, y =, y 3 =.5, y 4 =. S = x 3 (y + 4y + y + 4y 3 + y 4 ) =.5 ( ) = We have f (x) =, so the maximum value of f on the interval is, so the error estimate for Simpson s rule satisfies E S M(b a)5 8n 4 = ( ( )5 ) = The actual value of the integral (from above) is So the actual value of the error is So (x + ) = E S = (true value) (Simpson s value) = =. E S = %. (true value) (Simpson s rule gets the value exactly right!) 45

7 Section 7.6, Problem 7. (To make the notation more familiar, I ll use x instead of s as the variable, so we re doing the integral x.) Part I Trapezoidal Rule: We divide the interval from to into 4 pieces, so x = 4 =.5, and our x values are x =, x =.5, x =.5, x 3 =.75, x 4 =. The y values are (note y = f(x) = /x ) y =, y =.64, y =.4444, y 3 =.365, y 4 =.5. Then the trapezoidal rule gives T = x (y + y + y + y 3 + y 4 ) =.5 ( ) = We have f (x) = 6/x 4, so for x, the maximum of f (x) is M = 6, which occurs at x =. So the error estimate satisfies E T The actual value of the integral is M(b a)3 6( )3 n = 4 =.35. = x x So the actual value of the error is x= x= = =.5. E T = (true value) (trapezoidal value) = = So E T.8993 = =.8%. (true value).5 46

8 Part II Simpson s Rule: The first part is the same. We divide the interval from to into 4 pieces, so x =.5, and our x values are x =, x =.5, x =.5, x 3 =.75, x 4 =. The y values are (note y = f(x) = /x ) y =, y =.64, y =.4444, y 3 =.365, y 4 =.5. Then Simpson s rule gives S = x 3 (y + 4y + y + 4y 3 + y 4 ) =.5 ( ) 3 = The maximum of f (x) = 6/x 4 for x is M = f () = 6, so the error estimate for Simpson s rule satisfies E S M(b a)5 8n 4 = 6( )5 ) =.383. The actual value of the integral (from above) is So the actual value of the error is =.5. x E S = (true value) (Simpson s value) = =.476. So E S.476. == =.835%. (true value).5 47

9 Section 7.6, Problem 8(a). We want to use Simpson s rule to estimate e t dt. π We take n =, so t = / =. and the t values are t =, t =., t =.,... t 8 =.8, t 9 =.9, t =. We need to compute f(t) = e t for each of these values of t. This is easy to do on a calculator: t e t We then add up appropriate multiples as specified by Simpson s rule. So S = t ( f(t ) + 4f(t ) + f(t ) + 4f(t 3 ) + + f(t 8 ) + 4f(t 9 ) + f(t ) ) 3 = erf() = π e t dt π (If you re interested, erf() = to decimal places. The error is.9.) 48

10 Assignment 5 Assigned Fri Oct 8 Section 7.7, Problem. b x + = lim b x + = lim b tan (x) x=b = lim x= b tan (b) = π. Section 7.7, Problem 3. = lim x a + a x = lim a + x x= x=a = lim a + a =. Section 7.7, Problem 5. This integral has a potential problem at x =, which is in the middle of the interval, so we have to split the integral into two pieces. x = /3 = lim a x + /3 a x /3 + lim x/3 a + a x /3 = lim a 3x/3 x=a x= + lim a + 3x/3 x= x=a = lim a ( 3a /3 3( ) /3) + lim = ( + 3) + (3 ) = 6. a + ( 3 3a /3) Section 7.7, Problem 9. First we evaluate the indefinite integral using partial fractions. ( x = (x )(x + ) = x ) x + = ln x ln x + + C = ln x x + + C. Now we can compute x = lim a = lim a x= a x = lim ln x a x + x=a ( ) ln 3 ln a a + = ln 3 ln = ln 3. 49

11 Section 7.7, Problem. We first compute the indefinite integral using integration by parts, te t dt = te t e t dt u = t, du = dt, dv = e t dt, v = e t Then = te t e t + C = (t )e t + C. te t dt = lim a a te t dt = lim (t t= a )et = lim t=a a ( (a )e a ) =. (To compute the limit, use L Hôpital s rule, writing (a )e a as (a )/e a.) Section 7.7, Problem 39. This integral is actually easy to compute via substitution, x e /x = e u du u = /x, du = /x Then ln = e u + C x e /x = lim a + = e /x + C. ln a x e /x ln = lim e /x = lim e / ln e /a = e / ln. a + a a + (Notice that as a +, the quantity /a goes to, so e /a goes to.) Thus the integral converges. 5

12 Section 7.7, Problem 4. We have t + sin t t for all t π,, since sin t is positive for these values of t. Hence π dt t + sin t π Therefore the integral converges. dt π dt = lim t a a t = lim t t=π = lim π a = π. a t=a a Section 7.7, Problem 59. When x is big, e x is a lot bigger than x. We can make this precise, for example, e x > x for all x. (You can check this by looking at the graphs.) So e x x Therefore the integral diverges. x x x =. 5

13 Section 7.7, Problem 6. The function e x is a lot bigger than x, because e = is larger than. We can make this precise by noting that (( e x x = x e ) x ), and (e/) x.359 x.359 for all x. So we find that Taking the reciprocal gives Hence e x x.359 x. e x x.78 x. e x x.78 x. We can compute the integral x = x = e x ln = e x ln (ln ) + C = x (ln ) + C. So = lim x b b (ln ) + (ln ) = ln. Hence the original integral converges, because it is positive and smaller than a convergent integral. 5

14 Section 7.7, Problem 65. We first compute the indefinite integral du x(ln x) p = u p = p u p if p, ln u if p =, = p (ln x) p if p, ln ln x if p =. u = ln x, du = /x (a) x(ln x) p = lim a + a x(ln x) p lim = a + p (ln ) p p (ln a) p if p, lim ln ln ln ln a if p =. a + = p (ln ) p if p <, if p. So the integral converges if p < and diverges if p. (b) b x(ln x) p = lim b x(ln x) p lim = b p (ln b) p p (ln ) p if p, lim ln ln b ln ln if p =. b = p (ln ) p if p >, if p. So the integral converges if p > and diverges if p. 53