A =
Convert a decimal approximation to exact value symbolically
显示 更早的评论
Hi, I'm working with a definite integral and get a decimal approximation as my answer. I'd like to also get the exact solution:
syms x y
y1=sin(pi*x/2); y2=x; x1=-1; x2=1;
A=int((abs(y2-y1)),x1,x2);
Gives the answer as 0.2732 which is a correct decimal approximation. The exact solution is:
(4-pi)/pi
How do I get that?
采纳的回答
Here, you need the fact that y2-y1 is an odd function:
syms x y
y1=sin(pi*x/2); y2=x; x1=-1; x2=1;
A=2*int(y2-y1,x,x1,0)
6 个评论
rezheen
2025-7-8
If f is even or odd,
integral_{x=-1}^{x=1} abs(f(x)) dx = 2*integral_{x=-1}^{x=0} abs(f(x)) dx
Since in your case (y2-y1)(x) >=0 for -1 <= x <= 0, you get
integral_{x=-1}^{x=1} abs(y2(x)-y1(x)) dx = 2*integral_{x=-1}^{x=0} (y2(x)-y1(x)) dx
I don't know how MATLAB 2021a shows the result of this last integral. In R2024b, it's 4/pi - 1 .
I'm also not sure what even, odd functions have to do with spitting out the output.
If you don't make this simplification of splitting the integral, you won't get the symbolic output from above:
syms x y
y1=sin(pi*x/2); y2=x; x1=-1; x2=1;
A=int((abs(y2-y1)),x1,x2)
A=vpaintegral((abs(y2-y1)),x1,x2)
rezheen
2025-7-9
Walter Roberson
2025-7-10
If you have functions or areas that do not have identical halves, then it is unlikely that the function uses abs() [but possible...] int() tends to struggle a bit with abs()
Whether a function without abs() can be integrated or not... depends a lot on the function. It is easy to create functions that have no known exact integral.
Torsten
2025-7-10
if I have functions or areas without identical halves (odd?), I suppose there aren't exact solutions with the version I'm using.
Sometimes it works, sometimes not. The "abs" function is always problematic together with "int". And it doesn't have to do with your MATLAB version - it also doesn't work in R2024b as you can see above.
Huh, there appears to be a simplify bug.
syms x y
y1=sin(pi*x/2); y2=x; x1=-1; x2=1;
A=int((abs(y2-y1)),x1,x2)
simplify(A, 'steps', 50)
vpa(A)
更多回答（1 个）
Walter Roberson
2025-7-8
sympref('FloatingPointOutput',false);
Will display an unevaluated int() form for your original problem, and will display 4/pi - 1 for the version suggested by @Torsten
It seems that you have FloatPointOutput true in effect.
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