- Trapezium Rule

What is \(∫\sin⁡(x^3+x)\:dx\)?

After several attempts you will realize that this integral CANNOT be determined through traditional methods. However, if we needed to determine, for instance,\(\displaystyle ∫_0^{π\over2} \sin⁡(x^3+x)\:dx\) we can use a NUMERICAL METHOD. There are many numerical methods for determining a definite integral but we will be looking at the TRAPEZIUM RULE.

Here’s the derivation of the TRAPEZIUM RULE.
The area under the curve \(y=f(x)\) can be estimated by finding the sum of the areas of trapeziums of equal width (as shown).
The area of a trapezium with parallel sides \(y_0\) and \(y_1\) and width h is given by the formula \(\displaystyle {h\over2} (y_0+y_1)\).
The width of each trapezium is \(\displaystyle {b-a\over n}\) where \(n\) is the number of trapeziums.
Thus, we have the area under curve is
\(\displaystyle {h\over2} (y_0+y_1 )+{h\over2} (y_1+y_2 )+⋯+{h\over2} (y_{n-1}+y_n)\)
\(\displaystyle →{h\over2} (y_0+y_1+y_1+y_2+⋯+y_{n-1}+y_n)\)
\(\displaystyle →{h\over2} [(y_0+y_n)+2(y_1+y_2+⋯+y_{n-1})]\)
\(\displaystyle ∫_a^b \:y\: dx \approx {b-a \over 2n} [(y_0+y_n )+2(y_1+y_2+⋯+y_{n-1})]\)

VIDEO DERIVATION


LESSON 1
Using 5 trapeziums, estimate
\(\displaystyle ∫_1^3(x^2+1)\: dx\)
SOLUTION
  • Determine the width of each trapezium.
Width of each trapezium is \(\displaystyle {3-1\over5}=0.4\)
  • Determine the values \(y_0,y_1,…,y_n\).
\(y=f(x)=x^2+1\)
\(y_0=f(1.0)=(1.0)^2+1=2\)
\(y_1=f(1.4)=(1.4)^2+1=2.96\)
\(y_2=f(1.8)=(1.8)^2+1=4.24\)
\(y_3=f(2.2)=(2.2)^2+1=5.84\)
\(y_4=f(2.6)=(2.6)^2+1=7.76\)
\(y_5=f(3.0)=(3.0)^2+1=10\)
  • Substitute the values into the formula for the TRAPEZIUM RULE. Determine the answer to the required level of specificity i.e. decimal places or significant figures.
\(\displaystyle \int_a^b \:y \:dx \approx {b-a\over2n} [(y_0+y_n )+2(y_1+y_2+⋯+y_{n-1})]\)
\(\displaystyle ⟹\int_1^3 (x^2+1)\:dx\approx {3-1\over 2(5)} [(2+10)+2(2.96+4.24+5.84+7.76)]\)
≈10.72

LESSON 2
Use the trapezium rule, with ordinates at \(x=0, x=0.25, x=0.50, x=0.75\) and \(x=1\) to find an approximation to
\(\displaystyle ∫_0^1 {1\over 1+e^x} \:dx\)
SOLUTION
From the diagram we see that based on the 5 ordinates there are 4 trapeziums.
  • Determine the width of each trapezium.
Width of each trapezium is 0.25.
  • Determine the values \(y_0, y_1,…, y_n\).
\(\displaystyle y=f(x)={1\over 1+e^x}\)
\(\displaystyle y_0=f(0)={1\over 1+e^0}=0.5\)
\(\displaystyle y_1=f(0.25)={1\over 1+e^{0.25}}=0.4378\)
\(\displaystyle y_2=f(0.5)={1\over 1+e^{0.5}}=0.3775\)
\(\displaystyle y_3=f(0.75)={1\over 1+e^{0.75}}=0.3208\)
\(\displaystyle y_4=f(1)={1\over 1+e^1}=0.2689\)
  • Substitute the values into the formula for the TRAPEZIUM RULE. Determine the answer to the required level of specificity i.e. decimal places or significant figures.
\(\displaystyle ∫_a^b y\:dx \approx {b-a \over 2n} [(y_0+y_n )+2(y_1+y_2+⋯+y_{n-1})]\)
\(\displaystyle ⟹∫_0^1 {1\over 1+e^x}\:dx\approx {1-0 \over 2(4)} [(0.5+0.2689)+2(0.4378+0.3775+0.3208)]\)
\(\approx 0.3801\)

Discussion

0 comments