(a) Find the approximations T 8 and M 8 for the given integral. (Round your answ

Question

(a) Find the approximations T8 and M8 for the given integral. (Round your answer to six decimal places.) T8 =, M8 =

(b) Estimate the errors in the approximations T8 and M8 in part (a). (Use the fact that the range of the sine and cosine functions is bounded by ±1 to estimate the maximum error. Round your answer to seven decimal places.) |ET| , |EM|

(c) How large do we have to choose n so that the approximations Tn and Mn to the integral are accurate to within 0.0001? (Use the fact that the range of the sine and cosine functions is bounded by ±1 to estimate the maximum error.)

n for Tn

n for Mn

Explanation / Answer

For part a we will find Trapezoidal method where n= 8, a=0, b= 1

b-af(x)dxx2(f(x0)+2f(x1)+2f(x2)+...+2f(xn1)+f(xn)), where x=ba/n.

We have that a=0, b=1, n=8.

Therefore, x=10/8=1/8.

Divide interval [0,1] into n=8 subintervals of length x=1/8 with the following endpoints: a=0,1/8,1/4,3/8,1/2,5/8,3/4,7/8,1=b.

Now, we just evaluate function at those endpoints:

f(x0)=f(a)=f(0)=13=13

2f(x1)=2f(1/8)=26cos(1/64)=25.9968262364462

2f(x2)=2f(1/4)=26cos(1/16)=25.9492352782026

2f(x3)=2f(3/8)=26cos(9/64)=25.7433432976228

2f(x4)=2f(1/2)=26cos(1/4)=25.1917229644768

2f(x5)=2f(5/8)=26cos(25/64)=24.0414527981429

2f(x6)=2f(3/4)=26cos(9/16)=21.9940369800078

2f(x7)=2f(7/8)=26cos(49/64)=18.7446839045881

f(x8)=f(b)=f(1)=13cos(1)=7.02392997628582

Finally, just sum up above values and multiply by x/2=1/16: 1/16(13+25.9968262364462+25.9492352782026+...+18.7446839045881+7.02392997628582)=11.7303269647358

Answer: 11.7303270

Now for M8 it is mid point rule

Midpoint Sum (also Midpoint Approximation) uses midpoints of subinterval: b-af(x)dxx(f(x0+x1/2)+f(x1+x2/2)+f(x2+x3/2)+...+f(xn2+xn1/2)+f(xn1+xn/2)), where x=ba/n.

We have that a=0, b=1, n=8.

Therefore, x=10/8=1/8.

Divide interval [0,1] into n=8 subintervals of length x=1/8: a=[0,1/8],[1/8,1/4],[1/4,3/8],[3/8,1/2],[1/2,5/8],[5/8,3/4],[3/4,7/8],[7/8,1]=b.

Now, we just evaluate function at midpoints:

f(x0+x1/2)=f((0)+(1/8)/2)=f(1/16)=13cos(1/256)=12.9999008179972

f(x1+x2/2)=f((1/8)+(1/4)/2)=f(3/16)=13cos(9/256)=12.9919670749755

f(x2+x3/2)=f((1/4)+(3/8)/2)=f(5/16)=13cos(25/256)=12.9380604180719

f(x3+x4/2)=f((3/8)+(1/2)/2)=f(7/16)=13cos(49/256)=12.7625898583045

f(x4+x5/2)=f((1/2)+(5/8)/2)=f(9/16)=13cos(81/256)=12.3546768738464

f(x5+x6/2)=f((5/8)+(3/4)/2)=f(11/16)=13cos(121/256)=11.5747081464704

f(x6+x7/2)=f((3/4)+(7/8)/2)=f(13/16)=13cos(169/256)=10.2686534905003

f(x7+x8/2)=f((7/8)+(1)/2)=f(15/16)=13cos(22/5256)=8.29391886142123

Finally, just sum up above values and multiply by x=1/8: 1/8(12.9999008179972+12.9919670749755+12.9380604180719+...+10.2686534905003+8.29391886142123)=11.7730594426984

Answer: 11.7730595.

b) f(x) = 13 cos(x²)
f'(x) = -26x sin(x²)
f''(x) = -26sin(x²) - 26x * 2x cos(x²) = -26 (sin(x²) + 2x² cos(x²))

On interval 0 x 1
|f''(x)| 26 (1 + 2(1)) = 78

E(T) = -(b-a)³/(12N²) * f''(), for some between a and b
|E(T8)| 1/(12*8²) * 78
|E(T8)| 0.1015625

E(M) = (b-a)³/(24N²) * f''(), for some between a and b
|E(T8)| 1/(24*8²) * 78
|E(T8)| 0.05078125

C)Now we will use the result of b

From (b)

|E(Tn)| 1/(12n²) * 78 0.0001
6.5/n² 0.0001
6.5/0.0001 n²
n² 65000
n 254.95=255

n 255

|E(Mn)| 1/(24n²) * 78 0.0001
3.25/n² 0.0001
3.25/0.0001 n²
n² 32500
n 180.27

n 181 answers

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