(a) Estimate the area under the graph of f ( x ) = 2 + 4 x 2 from x = 1 to x = 2

Question

(a) Estimate the area under the graph of

f(x) = 2 + 4x2 from x = 1 to x = 2 using three rectangles and right endpoints.

R3 =  

Then improve your estimate by using six rectangles.
R6 =  

Sketch the curve and the approximating rectangles for R3.

Sketch the curve and the approximating rectangles for R6.

(b) Repeat part (a) using left endpoints.

Sketch the curve and the approximating rectangles for L3.

Sketch the curve and the approximating rectangles for L6.

(c) Repeat part (a) using midpoints.

Sketch the curve and the approximating rectangles for M3.

Sketch the curve and the approximating rectangles for M6.

(d) From your sketches in parts (a)-(c), which appears to be the best estimate?

L6M6    R6

Explanation / Answer

a)We have that a=1a=1, b=2b=2, n=3n=3.

Therefore, x=2(1)3=1x=2(1)3=1.

Divide interval [1,2][1,2] into n=3n=3 subintervals of length x=1x=1: a=[1,0],[0,1],[1,2]=ba=[1,0],[0,1],[1,2]=b.

Now, we just evaluate function at right endpoints:

f(x1)=f(0)=2=2f(x1)=f(0)=2=2

f(x2)=f(1)=6=6

f(x3)=f(b)=f(2)=18=18f(x3)=f(b)=f(2)=18=18

Finally, just sum up above values and multiply by x=1x=1: 1(2+6+18)=26

R3 = 26

Using 6 rectangles

We have that a=1a=1, b=2b=2, n=6n=6.

Therefore, x=2(1)6=12x=2(1)6=12.

Divide interval [1,2][1,2] into n=6n=6 subintervals of length x=12x=12: a=[1,12],[12,0],[0,12],[12,1],[1,32],[32,2]=ba=[1,12],[12,0],[0,12],[12,1],[1,32],[32,2]=b.

Now, we just evaluate function at right endpoints:

f(x1)=f(12)=3=3f(x1)=f(12)=3=3

f(x2)=f(0)=2=2f(x2)=f(0)=2=2

f(x3)=f(12)=3=3f(x3)=f(12)=3=3

f(x4)=f(1)=6=6f(x4)=f(1)=6=6

f(x5)=f(32)=11=11

f(x6)=f(b)=f(2)=18=18f(x6)=f(b)=f(2)=18=18

Finally, just sum up above values and multiply by x=12x=12: 12(3+2+3+...+11+18)=21.5

b)

We have that a=1a=1, b=2b=2, n=3n=3.

Therefore, x=2(1)3=1x=2(1)3=1.

Divide interval [1,2][1,2] into n=3n=3 subintervals of length x=1x=1: a=[1,0],[0,1],[1,2]=ba=[1,0],[0,1],[1,2]=b.

Now, we just evaluate function at left endpoints:

f(x0)=f(a)=f(1)=6=6f(x0)=f(a)=f(1)=6=6

f(x1)=f(0)=2=2

f(x2)=f(1)=6=6f(x2)=f(1)=6=6

Finally, just sum up above values and multiply by x=1x=1: 1(6+2+6)=14

Using 6 rectangles

We have that a=1a=1, b=2b=2, n=6n=6.

Therefore, x=2(1)6=12x=2(1)6=12.

Divide interval [1,2][1,2] into n=6n=6 subintervals of length x=12x=12: a=[1,12],[12,0],[0,12],[12,1],[1,32],[32,2]=ba=[1,12],[12,0],[0,12],[12,1],[1,32],[32,2]=b.

Now, we just evaluate function at left endpoints:

f(x0)=f(a)=f(1)=6=6f(x0)=f(a)=f(1)=6=6

f(x1)=f(12)=3=3f(x1)=f(12)=3=3

f(x2)=f(0)=2=2f(x2)=f(0)=2=2

f(x3)=f(12)=3=3f(x3)=f(12)=3=3

f(x4)=f(1)=6=6

f(x5)=f(32)=11=11f(x5)=f(32)=11=11

Finally, just sum up above values and multiply by x=12x=12: 12(6+3+2+...+6+11)=15.5

C)

We have that a=1a=1, b=2b=2, n=3n=3.

Therefore, x=2(1)3=1x=2(1)3=1.

Divide interval [1,2][1,2] into n=3n=3 subintervals of length x=1x=1: a=[1,0],[0,1],[1,2]=ba=[1,0],[0,1],[1,2]=b.

Now, we just evaluate function at midpoints:

f(x0+x12)=f((1)+(0)2)=f(12)=3=3f(x0+x12)=f((1)+(0)2)=f(12)=3=3

f(x1+x2/2)=f((0)+(1)2)=f(12)=3=3

f(x2+x3/2)=f((1)+(2)2)=f(32)=11=11f(x2+x32)=f((1)+(2)2)=f(32)=11=11

Finally, just sum up above values and multiply by x=1x=1: 1(3+3+11)=17

We have that a=1a=1, b=2b=2, n=6n=6.

Therefore, x=2(1)6=12x=2(1)6=12.

Divide interval [1,2][1,2] into n=6n=6 subintervals of length x=12x=12: a=[1,12],[12,0],[0,12],[12,1],[1,32],[32,2]=ba=[1,12],[12,0],[0,12],[12,1],[1,32],[32,2]=b.

Now, we just evaluate function at midpoints:

f(x0+x12)=f((1)+(12)2)=f(34)=174=4.25f(x0+x12)=f((1)+(12)2)=f(34)=174=4.25

f(x1+x2/2)=f((12)+(0)2)=f(14)=94=2.25f(x1+x22)=f((12)+(0)2)=f(14)=94=2.25

f(x2+x3/2)=f((0)+(12)2)=f(14)=94=2.25f(x2+x32)=f((0)+(12)2)=f(14)=94=2.25

f(x3+x4/2)=f((12)+(1)2)=f(34)=174=4.25f(x3+x42)=f((12)+(1)2)=f(34)=174=4.25

f(x4+x5/2)=f((1)+(32)2)=f(54)=334=8.25

f(x5+x6/2)=f((32)+(2)2)=f(74)=574=14.25f(x5+x62)=f((32)+(2)2)=f(74)=574=14.25

Finally, just sum up above values and multiply by x=12x=12: 12(4.25+2.25+2.25+...+8.25+14.25)=17.75

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