x2sin x, ∆x = π − 0 8 = π 8, (a) Use the Midpoint Rule, M8 = Z π 0 x2sin x dx ≈ π 8  f ( π 16

Chapter 7.7(Solution) 5. sol.

f (x) = x²sin x, ∆x = π − 0

8 = π

8, (a) Use the Midpoint Rule,

M₈ = Z ^π

0

x²sin x dx ≈ π 8

 f ( π

16) + f (3π

16) + · · · f (15π 16 )



≈ 5.9329566

(b) Use the Simpson’s Rule, S8 = Z ^π

0

x²sin x dx

≈ 1 3·π

8



f (0) + 4f (π

8) + 2f (2π

8 ) + 4f (3π

8 ) + 2f (4π

8 ) + 4f (5π

8 ) + 2f (6π

8 ) + 4f (7π

8 ) + f (π)



≈ 5.8692468 (c) Actual value:

Z ^π

0

x²sin x dx = −x²cos x  

π

0 + 2 Z ^π

0

x cos x dx

= π²+ 2



x sin x  

π

0 − Z π

0

sin x dx



= π²+ 2h 0 −

− cos x  

π

0

i

= π²− 4 ≈ 5.8696044

Error: EM = (π²− 4) − M8 ≈ −0.063352199; ES = (π²− 4) − S8 ≈ 0.000357601.

23. sol.

(a) f (x) = e^{cos x}, f^′(x) = e^{cos x}(− sin x), f^′′(x) = e^{cos x}(sin²x − cos x). From the graph of f^′′(x) (See Figure 1), we find that the maximum value of |f^′′(x)| occurs at the endpoint of the interval [0, 2π]. Because of f^′′(0) = −e, we choose K = e and

|f^′′(x)| ≤ K.

Figure 1: 7.7 Ex.23(a)

(b) ∆x = 2π − 0 10 = π

5, I = Z 2π

0

f (x) dx,

M10= ∆x

 f (π

10) + f (3π

10) + · · · + f (19π 10 )



≈ 7.954926518

1

(c) |E^M| ≤ K(b − a)³

24n² = e · (2π − 0)³

24 · 10² ≈ 0.280945995 (d) I =

Z ^π

0

x²sin x dx ≈ 7.954926521. (using software like CAS or Maple or ...) (e) The actual error is about 3 × 10⁻⁹; the estimate error in (c) is about 0.28.

So the actual error is much smaller.

(f) f^′′′(x) = e^{cos x}(− sin³x + 3 sin x cos x + sin x), and f⁽⁴⁾(x) = e^{cos x}(sin⁴x − 6 sin²x cos x − 7 sin²x + cos x + 3). From the graph of f⁽⁴⁾(x) (See Figure 2), we find that the maximum value of |f⁽⁴⁾(x)| occurs at the endpoint of the interval [0, 2π].

Because of f⁽⁴⁾(0) = 4e, we choose K = 4e and |f⁽⁴⁾(x)| ≤ K.

Figure 2: 7.7 Ex.23(f)

(g) ∆x = 2π − 0 10 = π

5, I = Z 2π

0

f (x) dx,

S10= ∆x 3



f (0) + 4f (π

5) + 2f (2π

5 ) + · · · + 4f (9π

5 ) + f (2π)



≈ 7.953789422

(h) |E^S| ≤ K(b − a)⁵

180n⁴ = 4e · (2π − 0)⁵

180 · 10⁴ ≈ 0.059153618

(i) The actual error is about 7.954926521 − 7.953789422 ≈ 0.001137099; the estimate error in (h) is about 0.059153618. So the actual error is smaller.

(j) We require |E^S| = K(b − a)⁵

180n⁴ ≤ 0.0001 ⇒ n⁴ ≥ K(b − a)⁵

180 · 0.0001 = 4e(2π)⁵ 180 · 0.0001 ≈ 5915361.766 ⇒ n⁴ ≥ 5915362 ⇒ n ≥ 49.3. So we require n ≥ 50 to guarantee that

|ES| ≤ 0.0001.

45. sol.

We divide the interval [a, b] into n intervals: [x_i−1, xi], for i = 1, 2, · · · , n when we use trapezoidal rule and midpoint rule, so we only consider one interval as shown in Figure 3. Let A, B, and E be x_i−1, xi, and xi; the red curve represents f (x) (f^′′(x) <

2

Figure 3: 7.7 Ex.45 0). The actual value ofR^b

af (x) dx is the area between the red curve and AB. Tn is the area of trapezoid AQRB, so Tⁿ < R^b

a f (x) dx. Mⁿ is the area of the rectangle with length AB and width EP , and it equals the area of the trapezoid ACDB, where CD is the tangent line of f (x) at point P . Therefore, Mn >R^b

af (x) dx.

46. sol.

Let f (x) be a polynomial of degree < 3, so assume f (x) = Ax³+ Bx²+ Cx + D.

Consider that we divide [a, b] into two subintervals (n = 2), if we prove that the estimate is exact for n = 2, then for the general case (n is a large even number), the estimate is the sum of n/2 exact values, so it is a exact value.

Assume that x0 = −h, x1 = 0, and x2 = h, the Simpson’s rule gives the approx- imation value:

Z ^h

−h

f (x) dx ≈ h

3[f (−h) + 4f (0) + f (h)]

= h

3(−Ah³+ Bh²− Ch + D) + 4D + (Ah³ + Bh²+ Ch + D)

= 2

3Bh³+ 2Dh.

The actual value of the integral is:

Z ^h

−h

f (x) dx = Z ^h

−h

(Ax³+ Bx²+ Cx + D) dx

= 1

4Ax⁴+1

3Bx³+ 1

2x² + Dx  

h

−h

= 2

3Bh³+ 2Dh.

47. sol.

Suppose we divide [a, b] into n subintervals with endpoints x0, x1, · · · xn (∆x = b − a

n ), and xi = 1

2(x_i−1+xi) is the midpoint of [x_i−1, xi] for i = 1, · · · , n. Therefore,

3

Mⁿ= ∆x ·

n

X

i=1

f (xⁱ),

Tⁿ = 1 2 · ∆x ·

"

f (x0) + 2

n−1

X

i=1

f (xⁱ) + f (xⁿ)

# ,

T_2n = 1

2 · ∆x 2

"

f (x₀) + 2

n−1

X

i=1

f (xi) + 2

n

X

i=1

f (xi) + f (xn)

#

Then, 1

2[Tn+ Mn] = T2n. 48. sol.

Suppose we divide [a, b] into n subintervals with endpoints x0, x1, · · · xn (∆x = b − a

n ), and xi = 1

2(x_i−1+xi) is the midpoint of [x_i−1, xi] for i = 1, · · · , n. Therefore, Mⁿ= ∆x ·

n

X

i=1

f (xⁱ),

Tⁿ = 1 2 · ∆x ·

"

f (x0) + 2

n−1

X

i=1

f (xⁱ) + f (xⁿ)

# , Then,

1

3Tn+ 2 3Mn

= 1

3[Tⁿ+ 2Mⁿ]

= 1 3

"

1

2· ∆x ·f (x0) + 2

n−1

X

i=1

f (xi) + f (xn)

!

+ 2 ∆x ·

n

X

i=1

f (xi)

!#

= 1

3 · ∆x 2

"

f (x0) + 2

n−1

X

i=1

f (xi) + f (xn) + 4

n

X

i=1

f (xi)

#

= S2n

4