# Area using trapezoidal rule

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Bibhudatta 2024년 4월 14일
댓글: Bibhudatta 2024년 4월 14일
I have been asked to calculate the area using trapezoidal rule for the following tabular data
x=[-2 0 2 4 8 8 10]
y=[35 5 -10 2 5 3 20]
I have written this function:
function[Area]=Trapz(x,y);
%Area is the total area i.e integration of the curve
%y=f(x)
%h=perpendicular distance between 2 parallel lines
h=x(2)-x(1);
n=length(y);%no of elements on the y axis
%sum of first and last y coordinate
sumy=y(1)+y(n);
remy=sum(y)-sumy;
Area=(1/2)*h*(sumy+2*remy)%Trapezoidal rule formula
end
My doubt is whether while calling the function, we should use the absolut values of y or not, as according to basic integral calculus, the area under a curve should not be negative.
If I use the absoulte values of y, I get area as 105, whereas if I do not take the absolute values of y, the are becomes 65. Which one is correct.
Help is appreciated.
Thanks

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### 채택된 답변

John D'Errico 2024년 4월 14일
편집: John D'Errico 2024년 4월 14일
Um, this is WRONG. Basic integral calculus does NOT say a result should never be negative. You are possibly misunderstanding what an integral means, as compared to the concept of area.
For example, what is the integral of the simple function y = -x, from 0 to 1? I'll use MATLAB here.
syms x
y(x) = -x;
PLOT IT FIRST. ALWAYS PLOT EVERYTHING.
fplot(y,[0,1])
xlabel X
ylabel Y
grid on
As you can see, y is negative over the entire region of interest.
int(y,0,1)
ans =
And in fact, the integral is -1/2. Yes. An integral CAN be negative, even though you may think of an integral as something representing area. It can be negative.
Let me ask a different question. What is the area between the x axis, and the function y, over the support for x of [0,1]? Now we would be using an integral to compute a POSITIVE area, because we are asking for an area between two lines. We might do it in a symbolic form:
ax(x) = sym(0);
int(ax - y,0,1)
ans =
Note that I've not needed to use abs at all there. If the curves actually cross, then I would need to think about absolute values to get the positive area.
The point here? IF you were asked to compute an area between things, then it makes sense that the result is positive. But if you are going to compute an integral, using a trapezoidal rule, then the sign is what it is. For example, what is the integral of the curve sin(x), for x between 0 and 2*pi?
fplot(sin(x),[0,2*pi])
yline(0)
int(sin(x),0,2*pi)
ans =
0
The two lobes cancel out, giving exactly zero. And trapezoidal rule would agree. I will just use trapz so I won't need to use (and validate) your code.
X = linspace(0,2*pi);
Y = sin(X);
trapz(X,Y)
ans = 3.9899e-17
Note tht trapz gives me some floating point crap, but it is still effectively zero.
BUT the area between the curve and the x axis is a different thing. That is known to be 4.
int(abs(0 - sin(x)),0,2*pi)
ans =
4
trapz(X,abs(0 - sin(X)))
ans = 3.9997
And to within a reasonable tolerance, trapz agrees. (Note that for such a function, a trapezoidal rule will always under-estimate the area.)
The problem is, there is an ambiguity, something we cannot resolve, because we don't see the problem statement. Did the problem statement explicitly request you compute the integral using trapezoidal rule? Did the problem statement ask you to compute an area between things? Or, is the problem statement itself a bit ambiguous, poorly stated, so it asks you to compute an area "under" a curve that is both positive and negative? In the latter case, only your teacher could remove the ambiguity of a poorly phrased problem. I would hope the question was well posed, and you have merely misinterpreted it, but teachers are sometimes sloppy in what they ask.
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Bibhudatta 2024년 4월 14일
Thanks John for the detailed answer. The exact question was as below:
Bibhudatta 2024년 4월 14일
From what you said, I understood that if we are asked to evaluate an integral using the trapezoidal rule we just have to estimate the integral value, no matter positive or negative. But if it is asked to find the area under/ between, we will need to consider the absolute values.
If this is true, I have misinterpreted the question.

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