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Trapezoidal Question - Help! (1 Viewer)
- Thread starter Lemiixem
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Nooblet94
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The function looks something like 
And since the definite integral is the area under the curve, you can find it by finding A1, A2, A3, A4 and adding them together:
Now, they're all trapezium's and the area of a trapezium is
)
(Where h is the height, a is the length of one side and b is the length of the other), so the expresion becomes
+\frac{h}{2}(a_2+b_2)+\frac{h}{2}(a_3+b_3)+\frac{h}{2}(a_4+b_4))
However, you may notice two things about that. Firstly, they all have the same height, so we can factorise the half h out. Secondly, they share sides such that
. Substituting those into the expression we end up with
)
Upon substituting in the values given in the table:
=75.1)
Obviously you wouldn't need to show that amount of working in an exam (You wouldn't need to factorise and whatnot), I just did that for fun
And since the definite integral is the area under the curve, you can find it by finding A1, A2, A3, A4 and adding them together:
Now, they're all trapezium's and the area of a trapezium is
(Where h is the height, a is the length of one side and b is the length of the other), so the expresion becomes
However, you may notice two things about that. Firstly, they all have the same height, so we can factorise the half h out. Secondly, they share sides such that
Upon substituting in the values given in the table:
Obviously you wouldn't need to show that amount of working in an exam (You wouldn't need to factorise and whatnot), I just did that for fun
Timske
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- Uni Grad
- 2016
Trapezoidal rule; A = h/2[y0 + yN + 2(y1+y2+y3+...+yN-1)]
y0= first term
yN= last term
X 1, 3, 5, 7, 9
f(x) 3.2, 5.9, 8.4, 11.6, 20.1
In this case h = 2
y0 = 3.2
y1 = 5.9
y2 = 8.4
y3 = 11.6
yN= 20.1
Simply plug those numbers in the Trapezoidal Rule so you get;
A=2/2[(3.2 + 20.1 + 2(5.9 + 8.4 + 11.6)]
=[23.2 +2(25.9)]
=23.3 + 51.8
=75.1 units
y0= first term
yN= last term
X 1, 3, 5, 7, 9
f(x) 3.2, 5.9, 8.4, 11.6, 20.1
In this case h = 2
y0 = 3.2
y1 = 5.9
y2 = 8.4
y3 = 11.6
yN= 20.1
Simply plug those numbers in the Trapezoidal Rule so you get;
A=2/2[(3.2 + 20.1 + 2(5.9 + 8.4 + 11.6)]
=[23.2 +2(25.9)]
=23.3 + 51.8
=75.1 units
[maths]Formula: [/maths] [maths]A=\frac{h}{2}(ends + 2(Middles))[/maths]
Table:
X 1, 3, 5, 7, 9
f(x) 3.2, 5.9, 8.4, 11.6, 20.1
h= Distance between (x) values which = 2
Ends of table of value are 3.2 and 20.1
Middle Values of table of value: are 5.9,8.4,11.6
Note its always like that end values of table and middle values
Know sub into formula
[maths]A=\frac{2}{2}[(3.2 + 20.1 + 2(5.9 + 8.4 + 11.6)] = 75.1units^2[/maths]
Remember you shouldnt use a = sign but instead use an approximation sign ≐
I know the guy above showed you but thought ill show my way might find it easier.
Table:
X 1, 3, 5, 7, 9
f(x) 3.2, 5.9, 8.4, 11.6, 20.1
h= Distance between (x) values which = 2
Ends of table of value are 3.2 and 20.1
Middle Values of table of value: are 5.9,8.4,11.6
Note its always like that end values of table and middle values
Know sub into formula
[maths]A=\frac{2}{2}[(3.2 + 20.1 + 2(5.9 + 8.4 + 11.6)] = 75.1units^2[/maths]
Remember you shouldnt use a = sign but instead use an approximation sign ≐
I know the guy above showed you but thought ill show my way might find it easier.
Last edited: