#$&*
course Mth 174
Section 7.61) 7.6.1 Estimate Int(x^2 dx,0,6) using SIMP(2).
SIMP(2) = [2MID(2) + TRAP(2)]/3 = [2f(1.5)(3) + 2f(4.5)(3) + 3(f(0)+f(3))/2 + 3(f(3) + f(6))/2]/3 = 72
@& Good.
Note the following:
For linear functions the trapezoidal and midpoint approximations are both completely accurate (since the graph is a straight line, the two are in fact identical), and the Simpson approximation is therefore completely accurate.
For a quadratic function which is concave up, the trapezoidal rule will overestimate and the midpoint approximation will underestimate, the underestimate being half the overrestimate. This is a unique property of quadratic functions, and makes the Simpson approximation completely accurate for such functions. For quadratic functions which are concave down, an analogous statement holds, so that the Simpson approximation is exact for all quadratic functions.*@
2) 7.6.3 Given the integral Int(1/x dx, 1, 2)
• Using n = 2,4,8,16,32,64,128 find the left and right approximations and the corresponding errors.
• What are the signs of the erros in the left and right approximations? How do the errors change when n is doubled?
• For the same values of n, compute the midpoint and trapezoidal approximations and corresponding errors.
• What are the signs of the errors for the midpoint and trapezoid approximations? How do the errors change if n is doubled?
• For n = 2,4,8,16,32 compute Simpson's rule approx. and the error for each n. How do these errors change as you double n?
N actual value left error of left right error of right
2 0.69314718
4 0.69314718
8 0.69314718
16 0.69314718
32 0.69314718
64 0.69314718
128 0.69314718
3) 7.6.6 The approximation of a definite integral using n = 10 is 2.346 and the exact value is 4.0. If the approximation was found using each of the following rules , use the same rule to estimate the integral with n = 30. LEFT, TRAP, and SIMP.
LEFT(30) =4.0 - 10/30 (4 - 2.346) = 3.449
TRAP(30) = 4.0 - (10/30)^2 (4 - 2.346) = 3.816
SIMP(30) = 4.0 - (10/30)^4 (4 - 2.346) = 3.980
4) Let a < m < b, with m the midpoint between a and b.
• If f(x) = x^2, show that int(f(x), x, a, b) = h/3 ( f(a)/2 + 2 f(m) + f(b) / 2).
@& f(a) / 2 = a^2 / 2.
What are the expressions for f(m) and f(b)?
What therefore is the expression for
h/3 ( f(a)/2 + 2 f(m) + f(b) / 2).
Note that h here means the length of the interval, which is b - a.
Simplify your expression and see if it is equal to integral( x^2 dx, x from a to b).
*@
• If f(x) is amy quadratic function, show that int(f(x), x, a, b) = h/3 ( f(a)/2 + 2 f(m) + f(b) / 2).
Explain how this proves that Simpson's Rule gives the exact value of the integral when f(x) is a quadratic function.
5) If TRAP(10) = 12.676 and TRAP(30) = 10.420, estimate the actual value of the integral.
TRAP(30) = A.V - (10/30)^2 (A.V - TRAP(10))
A.V - 10.240 = (A.V - 12.676)/9
A.V = 9.936
=====
Section 7.7
1) 7.7.6 Calculate the integral Int(1/(x+4)^2 dx, 3, inf) if it converges. (Note: inf here means infinity)
Int(1/(x+4)^2 dx, 3, inf) = int((x+4)^-2 dx, 3, b) = (-(x+4)^-1, 3 , b) = -1/(b+4) + 1/7
As b approach inf, Int(1/(x+4)^2 dx, 3, inf) = 1/7. Therefore, it is converges.
2) 7.7.12 Calculate the integral Int(1/(u^2 - 16) dx, 0, 4) if it converges.
Int(1/(u^2 - 16) dx, 0, 4) = -ln|(y-4)/(y+4)|/8 0, 4 =lim(b to 0+ )= -ln(b)/8 + ln1/8 = inf. it is not converges
3) 7.7.24 Calculate the integral Int(3/(x ln x) dx, 1 2) if it converges.
Int(3/(x ln x) dx, 1 2) = lim(a to 2) int(3/(x ln x) dx, 1 2) = 3ln|lna| - 3ln|ln1
therefore, Int(3/(x ln x) dx, 1 2) is diverges.
4) 7.7.30 Calculate the integral Int(100/sqrt(y-6) dy, 7, inf) if it converges.
Int(100/sqrt(y-6) dy, 7, inf) = 100 lim(a to inf) int((y-6)^(-1/2) dy) = 100(2(y-6)^(1/2) 7, a) = 200sqrt(a-7) - 200sqrt(1)
Int(100/sqrt(y-6) dy, 7, inf) does not converge
5) 7.7.36 For what values of p does the integral Int(x^p ln x dx, 0, e) converge? What is the value of the integral when it does converge?
Int(x^p ln x dx, 0, e) = 1/(p+1) x^(p+1) lnx - 1/(p+1)^2 x^(p+1) + C = e^(p+1)/(p+1) - e^(p+1)/(p+1)^2 = pe^(p+1)/(p+1)^2
if p = 1 then, this is not converges.
@& That would be p = -1.
However you need to consider the fact that ln(x) is not defined at x = 0, and in fact approaches -infinity as x -> 0.
What do you think happens?
Note that you can consider how the value of x^(p + 1) ln(x) changes as x -> 0. The higher the power the more quickly x^(p+1) approaches 0, and if that term approaches 0 faster than ln(x) the integral can still converge.*@
6) 7.7.40 The kth moment mk of the normal distribution is given by mk = 1/sqrt(2pi) * Int( x^k * e ^ (-x^2/2) dx, -inf, inf). Use the fact that Int(e^(-x^2/2) dx, -inf, inf) = sqrt(2pi) to find m2 and m4.
@& You can do one or more integrations by parts to prove this result.*@
7) 7.7.44 The rate, r, at which people get sick during an epidemic of the flu can be modeled by r = 1000te^(-0.5t) where r is people/day and t is days since start of epidemic.
• Sketch and describe the graph of r as a function of t.
• At what time are people getting sick the fastest?
• How many people get sick altogether?
the graph is increasing first and then decreasing and approach to y = 0 as x to infinite.
dr/dt = 1000e^(-0.5t) - 500 te^(-0.5t) = 500(2-t)e^(-0.5t) = 0, then t =2
r has its maximum at x =2, where r = 735.759
int(1000te^(-0.5t) dt, 0, inf) = 1000[-2te^(-0.5t) + 2int(e^-0.5t) 0, inf]= 1000(-2te^(-0.5t) - 4e^(-0.5t) o, inf) = 4000e^(0) = 4000
@& Good.*@
@& See my notes and see if you can prove / solve some of the problems you didn't address.*@