Dr. J's Maths.com
Where the techniques of Maths
are explained in simple terms.

Calculus - Integration - Approximation methods.
Trapezoidal Rule - Test Yourself 1.


 

 

1.

  2.

Note there are two triangles with which we will use the normal "area of a triangle" formula.

The green triangle is however below the x-axis with a negative y value for its height. Calculating the area therefore gives a negative value.

As Area cannot be negative, we take the absolute value before adding:

  3.
  4. (i)

(ii)

(iv) Area = 1×1.25 + 1×3.25 = 4.5 u2.

(v) Comparing the areas between the curve and the respective areas in the rectangles shows the estimate using the rectangles is very close to the actual area under the curve. By observation, the area under the rectangles is slightly less than the total area under the curve.

3 function values

5. (i) y = 2x - x2 with the area bounded by the curve and the x-axis
where y ≥ 0 shaded.

(iii)

x 0 1 2
2x - x2 0 1 0

(iv) To determine if the value calculated using the Trapezoidal Rule is an over- or under-estimate, always draw the tops of each trapezium between their relevant parallel lines. The diagram below show this.

The black lines are under the outline of the curve. Hence the Trapezoidal estimate is an under-estimate of the actual area.

 

  6.
x 0 2 4
2x 1 4 16

  7.
x e 2e 3e
f(x) = logex 1 1.693 2.099

  8.
x 1 4 7
1 + 2 log10 x 1 2.204 2.690

  9.
x π/3 π/2 2π/3
sin x 0.866 1 0.866

More than 3 function values. 10. (i)
x 2 2.5 3 3.5 4
f(x) 2.773 5.727 9.888 15.346 22.181

(ii)

  11. Find an approximation for .
x 1 1.5 2 2.5 3
f(x) 11.2 17.8 9.3 4.1 11.6

  12. (i)
Spacing 0 100 200 300 400
Depth 40 80 50 70 30

(ii) Volume = 23,500m2 × 200 m

= 4,700,000 m3 = 4.7 million cubic metres.

 

13. (i)

(ii)

x π/3 π/2 2π/3 5π/6 π
sin 2x 0.866 0 -0.866 -0.866 0

As the sin 2x curve goes beneath the x axis and we need to calculate the area NOT evaluate the integral, we need to do the calculation in two parts - from π/3 to π/2 and then from π/2 to π. Then reverse the sign of the 2nd calculation before adding.

  14. Use 4 sub-intervals to find an approximation to .
x 0 1 2 3 4
f(x) 3 1.810 0.633 0.337 0.488

Diagram

15.

 

(i)

Time 0 1 2 3
Velocity 0.8 15 18 20

Distance = 0.5×(15.8 + 33 + 38) = 634.9 m.

(ii) The calculated distance would be slightly less than the actual distance travelled. The straight lines joining successive dots to make the trapezia would all be under the actual curve although reasonable close to it.

 

16.

 

0 5 10 15 20
0 8 12 10 0

Adding across the 2nd row to get sums of successive pairs:

Area = 2.5×(8 + 20 + 22 + 10) = 150 m2.

 

17.

(i)

0 5 10 15 20
1.5 3.2 4.2 5.6 2.4

(ii) Volume = 74.75 × 2 m × 60 min = 8,970 m3/hour

= 8,970 Ml/hour.

 

18.

  1 2 3
y = ln(x + 1) 0.693 1.099 1.386
0 -0.50 -0.667
Difference 0.693 1.599 2.053