is there any way i can integrate function of two variable, say
有什么方法可以让两个变量的函数积分?
f=@(x) x^2 + x*y
f = @(x)x ^ 2 + y *
over just x
在x
tried quad(f, a, b)
试四(f,a,b)
but doesn't work, looking for alternative solution
但不工作，寻找替代解决方案。
2 个解决方案
#1
0
Sorry, but quad does not solve symbolic problems. It does only numerical integration.
抱歉，但quad不能解决符号问题。它只做数值积分。
syms x y
int(x^2 + x*y,x)
ans =
(x^2*(2*x + 3*y))/6
The natural way to solve a symbolic problem is to use a symbolic tool.
解决符号问题的自然方法是使用符号工具。
From the followup, Anya wants something in-between. To steal the words of an old rock star named Mick, "You can't always get what you want."
接下来，安雅想要一些介于两者之间的东西。要偷一个叫米克的老摇滚明星的话，“你不能总是得到你想要的。”
Again, quad can't be used if you wish to integrate ONLY over x, as quad is an adaptive tool.
同样，如果您希望仅对x进行集成，则quad不能被使用，因为quad是一种自适应工具。
In SOME SIMPLE cases, you can use a simple tool like Simpson's rule to do the work. For example, suppose you wanted to solve the above problem, with x integrated over the interval [0 1]. For purposes of comparison, I'll do it symbolically first.
在一些简单的情况下，您可以使用像Simpson规则这样的简单工具来完成工作。例如，假设您想要解决上述问题，在区间[0 1]中集成了x。为了便于比较，我先象征性地做一下。
syms x y
res = int(x^2 + x*y,x);
subs(res,x,1) - subs(res,0)
ans =
y/2 + 1/3
Now, lets try it using a numerical integration on x.
现在，让我们用x的数值积分来试试。
syms y
x = 0:.01:1;
coef = mod((0:100)',2)*2 + 2;
coef([1 end]) = 1;
coef = 0.01*coef/3;
(x.^2 + x.*y)*coef
ans =
y/2 + 1/3
So in this very SIMPLE case, it did work. How about something a little more complicated? Integrate x*exp(x*y) over the interval [-1 1]. Again, a known form is accessible symbolically.
在这个非常简单的例子中，它确实起作用了。稍微复杂点的怎么样?积分x*exp(x*y)除以区间[-1 -1]。同样，一个已知的形式可以被象征性地访问。
syms x y
res = int(x*exp(x*y),x);
res = subs(res,x,1) - subs(res,-1)
res =
(exp(-y)*(y + 1))/y^2 + (exp(y)*(y - 1))/y^2
To test it out later, what value does this take on at y = 1/2?
为了检验它，y = 1/2的值是多少?
vpa(subs(res,y,1/2))
ans =
0.34174141687554424792549563431876
Lets try the same trick, using Simpson's rule.
让我们用辛普森法则试试同样的方法。
syms y
x = -1:.01:1;
coef = mod((-100:100)',2)*2 + 2;
coef([1 end]) = 1;
coef = 0.01*coef/3;
res = (x.*exp(x*y))*coef
res =
exp(y/2)/300 - exp(-y/2)/300 - exp(-y)/300 - exp(-y/4)/300 + exp(y/4)/300 - exp(-y/5)/750 + exp(y/5)/750 - exp(-(3*y)/4)/100 - exp(-(2*y)/5)/375 + exp((2*y)/5)/375 + exp((3*y)/4)/100 - exp(-(3*y)/5)/250 + exp((3*y)/5)/250 - (2*exp(-(4*y)/5))/375 + (2*exp((4*y)/5))/375 - exp(-y/10)/1500 + exp(y/10)/1500 - exp(-(3*y)/10)/500 + exp((3*y)/10)/500 - (7*exp(-(7*y)/10))/1500 + (7*exp((7*y)/10))/1500 - (3*exp(-(9*y)/10))/500 + (3*exp((9*y)/10))/500 - exp(-y/20)/1500 + exp(y/20)/1500 - exp(-(3*y)/20)/500 + exp((3*y)/20)/500 - exp(-y/25)/3750 + exp(y/25)/3750 - (7*exp(-(7*y)/20))/1500 - exp(-(2*y)/25)/1875 + exp((2*y)/25)/1875 + (7*exp((7*y)/20))/1500 - exp(-(3*y)/25)/1250 + exp((3*y)/25)/1250 - (3*exp(-(9*y)/20))/500 - (2*exp(-(4*y)/25))/1875 + (2*exp((4*y)/25))/1875 + (3*exp((9*y)/20))/500 - (11*exp(-(11*y)/20))/1500 - exp(-(6*y)/25)/625 + exp((6*y)/25)/625 + (11*exp((11*y)/20))/1500 - (7*exp(-(7*y)/25))/3750 + (7*exp((7*y)/25))/3750 - (13*exp(-(13*y)/20))/1500 - (4*exp(-(8*y)/25))/1875 + (4*exp((8*y)/25))/1875 + (13*exp((13*y)/20))/1500 - (3*exp(-(9*y)/25))/1250 + (3*exp((9*y)/25))/1250 - (11*exp(-(11*y)/25))/3750 + (11*exp((11*y)/25))/3750 - (17*exp(-(17*y)/20))/1500 - (2*exp(-(12*y)/25))/625 + (2*exp((12*y)/25))/625 + (17*exp((17*y)/20))/1500 - (13*exp(-(13*y)/25))/3750 + (13*exp((13*y)/25))/3750 - (19*exp(-(19*y)/20))/1500 - (7*exp(-(14*y)/25))/1875 + (7*exp((14*y)/25))/1875 + (19*exp((19*y)/20))/1500 - (8*exp(-(16*y)/25))/1875 + (8*exp((16*y)/25))/1875 - (17*exp(-(17*y)/25))/3750 + (17*exp((17*y)/25))/3750 - (3*exp(-(18*y)/25))/625 + (3*exp((18*y)/25))/625 - (19*exp(-(19*y)/25))/3750 + (19*exp((19*y)/25))/3750 - (7*exp(-(21*y)/25))/1250 + (7*exp((21*y)/25))/1250 - (11*exp(-(22*y)/25))/1875 + (11*exp((22*y)/25))/1875 - (23*exp(-(23*y)/25))/3750 + (23*exp((23*y)/25))/3750 - (4*exp(-(24*y)/25))/625 + (4*exp((24*y)/25))/625 - exp(-y/50)/7500 + exp(y/50)/7500 - exp(-(3*y)/50)/2500 + exp((3*y)/50)/2500 - (7*exp(-(7*y)/50))/7500 + (7*exp((7*y)/50))/7500 - (3*exp(-(9*y)/50))/2500 + (3*exp((9*y)/50))/2500 - (11*exp(-(11*y)/50))/7500 + (11*exp((11*y)/50))/7500 - (13*exp(-(13*y)/50))/7500 + (13*exp((13*y)/50))/7500 - (17*exp(-(17*y)/50))/7500 + (17*exp((17*y)/50))/7500 - (19*exp(-(19*y)/50))/7500 + (19*exp((19*y)/50))/7500 - (7*exp(-(21*y)/50))/2500 + (7*exp((21*y)/50))/2500 - (23*exp(-(23*y)/50))/7500 + (23*exp((23*y)/50))/7500 - (9*exp(-(27*y)/50))/2500 + (9*exp((27*y)/50))/2500 - (29*exp(-(29*y)/50))/7500 + (29*exp((29*y)/50))/7500 - (31*exp(-(31*y)/50))/7500 + (31*exp((31*y)/50))/7500 - (11*exp(-(33*y)/50))/2500 + (11*exp((33*y)/50))/2500 - (37*exp(-(37*y)/50))/7500 + (37*exp((37*y)/50))/7500 - (13*exp(-(39*y)/50))/2500 + (13*exp((39*y)/50))/2500 - (41*exp(-(41*y)/50))/7500 + (41*exp((41*y)/50))/7500 - (43*exp(-(43*y)/50))/7500 + (43*exp((43*y)/50))/7500 - (47*exp(-(47*y)/50))/7500 + (47*exp((47*y)/50))/7500 - (49*exp(-(49*y)/50))/7500 + (49*exp((49*y)/50))/7500 - exp(-y/100)/7500 + exp(y/100)/7500 - exp(-(3*y)/100)/2500 + exp((3*y)/100)/2500 - (7*exp(-(7*y)/100))/7500 + (7*exp((7*y)/100))/7500 - (3*exp(-(9*y)/100))/2500 + (3*exp((9*y)/100))/2500 - (11*exp(-(11*y)/100))/7500 + (11*exp((11*y)/100))/7500 - (13*exp(-(13*y)/100))/7500 + (13*exp((13*y)/100))/7500 - (17*exp(-(17*y)/100))/7500 + (17*exp((17*y)/100))/7500 - (19*exp(-(19*y)/100))/7500 + (19*exp((19*y)/100))/7500 - (7*exp(-(21*y)/100))/2500 + (7*exp((21*y)/100))/2500 - (23*exp(-(23*y)/100))/7500 + (23*exp((23*y)/100))/7500 - (9*exp(-(27*y)/100))/2500 + (9*exp((27*y)/100))/2500 - (29*exp(-(29*y)/100))/7500 + (29*exp((29*y)/100))/7500 - (31*exp(-(31*y)/100))/7500 + (31*exp((31*y)/100))/7500 - (11*exp(-(33*y)/100))/2500 + (11*exp((33*y)/100))/2500 - (37*exp(-(37*y)/100))/7500 + (37*exp((37*y)/100))/7500 - (13*exp(-(39*y)/100))/2500 + (13*exp((39*y)/100))/2500 - (41*exp(-(41*y)/100))/7500 + (41*exp((41*y)/100))/7500 - (43*exp(-(43*y)/100))/7500 + (43*exp((43*y)/100))/7500 - (47*exp(-(47*y)/100))/7500 + (47*exp((47*y)/100))/7500 - (49*exp(-(49*y)/100))/7500 + (49*exp((49*y)/100))/7500 - (17*exp(-(51*y)/100))/2500 + (17*exp((51*y)/100))/2500 - (53*exp(-(53*y)/100))/7500 + (53*exp((53*y)/100))/7500 - (19*exp(-(57*y)/100))/2500 + (19*exp((57*y)/100))/2500 - (59*exp(-(59*y)/100))/7500 + (59*exp((59*y)/100))/7500 - (61*exp(-(61*y)/100))/7500 + (61*exp((61*y)/100))/7500 - (21*exp(-(63*y)/100))/2500 + (21*exp((63*y)/100))/2500 - (67*exp(-(67*y)/100))/7500 + (67*exp((67*y)/100))/7500 - (23*exp(-(69*y)/100))/2500 + (23*exp((69*y)/100))/2500 - (71*exp(-(71*y)/100))/7500 + (71*exp((71*y)/100))/7500 - (73*exp(-(73*y)/100))/7500 + (73*exp((73*y)/100))/7500 - (77*exp(-(77*y)/100))/7500 + (77*exp((77*y)/100))/7500 - (79*exp(-(79*y)/100))/7500 + (79*exp((79*y)/100))/7500 - (27*exp(-(81*y)/100))/2500 + (27*exp((81*y)/100))/2500 - (83*exp(-(83*y)/100))/7500 + (83*exp((83*y)/100))/7500 - (29*exp(-(87*y)/100))/2500 + (29*exp((87*y)/100))/2500 - (89*exp(-(89*y)/100))/7500 + (89*exp((89*y)/100))/7500 - (91*exp(-(91*y)/100))/7500 + (91*exp((91*y)/100))/7500 - (31*exp(-(93*y)/100))/2500 + (31*exp((93*y)/100))/2500 - (97*exp(-(97*y)/100))/7500 + (97*exp((97*y)/100))/7500 - (33*exp(-(99*y)/100))/2500 + (33*exp((99*y)/100))/2500 + exp(y)/300
So I got a result, but its not the analytical one I wanted, and a bit of a nasty mess. Is it correct?
所以我得到了一个结果，但它不是我想要的分析结果，还有一点糟糕的混乱。是正确的吗?
vpa(subs(res,y,1/2))
ans =
0.34174141693463006644516447861307
I'll copy the analytical result from above so we can compare...
我将从上面复制分析结果以便我们可以比较……
0.34174141687554424792549563431876
As you can see, Simpson's rule, at a step size of 0.01 over [-1,1], did reasonably well, agreeing out to about 9 decimal digits.
正如你所看到的，辛普森的规则，步长为0.01 /(-1,1)，做得相当好，同意约9位小数。
There is no assurance that this technique will work as well on any more general kernel, but it might give you what you desire.
并不能保证这种技术能在任何更通用的内核中发挥作用，但是它可能会给您想要的东西。
#2
1
Looks like you want something like this:
看起来你想要这样的东西:
y = 100; % whatever y is
a = 0;
b = 2;
% you'll need to vectorize the integrand function
f = @(x) x.*x + x.*y
val = quad(f, a, b);
However, if you are looking for an algebraic answer, you'll need to use the Symbolic Toolbox, or some other software, or your calculus book. :-)
但是，如果您正在寻找一个代数的答案，您将需要使用符号工具箱，或者其他一些软件，或者您的微积分书。:-)
The whole "vectorize" thing comes from the Mathworks quad documentation that says:
整个“vectorize”是来自Mathworks quad文档，它说:
The function y = fun(x) should accept a vector argument x and return a vector result y, the integrand evaluated at each element of x.
函数y = fun(x)应该接受一个向量参数x，并返回一个向量结果y，并在x的每个元素上进行积分。
#1
0
Sorry, but quad does not solve symbolic problems. It does only numerical integration.
抱歉，但quad不能解决符号问题。它只做数值积分。
syms x y
int(x^2 + x*y,x)
ans =
(x^2*(2*x + 3*y))/6
The natural way to solve a symbolic problem is to use a symbolic tool.
解决符号问题的自然方法是使用符号工具。
From the followup, Anya wants something in-between. To steal the words of an old rock star named Mick, "You can't always get what you want."
接下来，安雅想要一些介于两者之间的东西。要偷一个叫米克的老摇滚明星的话，“你不能总是得到你想要的。”
Again, quad can't be used if you wish to integrate ONLY over x, as quad is an adaptive tool.
同样，如果您希望仅对x进行集成，则quad不能被使用，因为quad是一种自适应工具。
In SOME SIMPLE cases, you can use a simple tool like Simpson's rule to do the work. For example, suppose you wanted to solve the above problem, with x integrated over the interval [0 1]. For purposes of comparison, I'll do it symbolically first.
在一些简单的情况下，您可以使用像Simpson规则这样的简单工具来完成工作。例如，假设您想要解决上述问题，在区间[0 1]中集成了x。为了便于比较，我先象征性地做一下。
syms x y
res = int(x^2 + x*y,x);
subs(res,x,1) - subs(res,0)
ans =
y/2 + 1/3
Now, lets try it using a numerical integration on x.
现在，让我们用x的数值积分来试试。
syms y
x = 0:.01:1;
coef = mod((0:100)',2)*2 + 2;
coef([1 end]) = 1;
coef = 0.01*coef/3;
(x.^2 + x.*y)*coef
ans =
y/2 + 1/3
So in this very SIMPLE case, it did work. How about something a little more complicated? Integrate x*exp(x*y) over the interval [-1 1]. Again, a known form is accessible symbolically.
在这个非常简单的例子中，它确实起作用了。稍微复杂点的怎么样?积分x*exp(x*y)除以区间[-1 -1]。同样，一个已知的形式可以被象征性地访问。
syms x y
res = int(x*exp(x*y),x);
res = subs(res,x,1) - subs(res,-1)
res =
(exp(-y)*(y + 1))/y^2 + (exp(y)*(y - 1))/y^2
To test it out later, what value does this take on at y = 1/2?
为了检验它，y = 1/2的值是多少?
vpa(subs(res,y,1/2))
ans =
0.34174141687554424792549563431876
Lets try the same trick, using Simpson's rule.
让我们用辛普森法则试试同样的方法。
syms y
x = -1:.01:1;
coef = mod((-100:100)',2)*2 + 2;
coef([1 end]) = 1;
coef = 0.01*coef/3;
res = (x.*exp(x*y))*coef
res =
exp(y/2)/300 - exp(-y/2)/300 - exp(-y)/300 - exp(-y/4)/300 + exp(y/4)/300 - exp(-y/5)/750 + exp(y/5)/750 - exp(-(3*y)/4)/100 - exp(-(2*y)/5)/375 + exp((2*y)/5)/375 + exp((3*y)/4)/100 - exp(-(3*y)/5)/250 + exp((3*y)/5)/250 - (2*exp(-(4*y)/5))/375 + (2*exp((4*y)/5))/375 - exp(-y/10)/1500 + exp(y/10)/1500 - exp(-(3*y)/10)/500 + exp((3*y)/10)/500 - (7*exp(-(7*y)/10))/1500 + (7*exp((7*y)/10))/1500 - (3*exp(-(9*y)/10))/500 + (3*exp((9*y)/10))/500 - exp(-y/20)/1500 + exp(y/20)/1500 - exp(-(3*y)/20)/500 + exp((3*y)/20)/500 - exp(-y/25)/3750 + exp(y/25)/3750 - (7*exp(-(7*y)/20))/1500 - exp(-(2*y)/25)/1875 + exp((2*y)/25)/1875 + (7*exp((7*y)/20))/1500 - exp(-(3*y)/25)/1250 + exp((3*y)/25)/1250 - (3*exp(-(9*y)/20))/500 - (2*exp(-(4*y)/25))/1875 + (2*exp((4*y)/25))/1875 + (3*exp((9*y)/20))/500 - (11*exp(-(11*y)/20))/1500 - exp(-(6*y)/25)/625 + exp((6*y)/25)/625 + (11*exp((11*y)/20))/1500 - (7*exp(-(7*y)/25))/3750 + (7*exp((7*y)/25))/3750 - (13*exp(-(13*y)/20))/1500 - (4*exp(-(8*y)/25))/1875 + (4*exp((8*y)/25))/1875 + (13*exp((13*y)/20))/1500 - (3*exp(-(9*y)/25))/1250 + (3*exp((9*y)/25))/1250 - (11*exp(-(11*y)/25))/3750 + (11*exp((11*y)/25))/3750 - (17*exp(-(17*y)/20))/1500 - (2*exp(-(12*y)/25))/625 + (2*exp((12*y)/25))/625 + (17*exp((17*y)/20))/1500 - (13*exp(-(13*y)/25))/3750 + (13*exp((13*y)/25))/3750 - (19*exp(-(19*y)/20))/1500 - (7*exp(-(14*y)/25))/1875 + (7*exp((14*y)/25))/1875 + (19*exp((19*y)/20))/1500 - (8*exp(-(16*y)/25))/1875 + (8*exp((16*y)/25))/1875 - (17*exp(-(17*y)/25))/3750 + (17*exp((17*y)/25))/3750 - (3*exp(-(18*y)/25))/625 + (3*exp((18*y)/25))/625 - (19*exp(-(19*y)/25))/3750 + (19*exp((19*y)/25))/3750 - (7*exp(-(21*y)/25))/1250 + (7*exp((21*y)/25))/1250 - (11*exp(-(22*y)/25))/1875 + (11*exp((22*y)/25))/1875 - (23*exp(-(23*y)/25))/3750 + (23*exp((23*y)/25))/3750 - (4*exp(-(24*y)/25))/625 + (4*exp((24*y)/25))/625 - exp(-y/50)/7500 + exp(y/50)/7500 - exp(-(3*y)/50)/2500 + exp((3*y)/50)/2500 - (7*exp(-(7*y)/50))/7500 + (7*exp((7*y)/50))/7500 - (3*exp(-(9*y)/50))/2500 + (3*exp((9*y)/50))/2500 - (11*exp(-(11*y)/50))/7500 + (11*exp((11*y)/50))/7500 - (13*exp(-(13*y)/50))/7500 + (13*exp((13*y)/50))/7500 - (17*exp(-(17*y)/50))/7500 + (17*exp((17*y)/50))/7500 - (19*exp(-(19*y)/50))/7500 + (19*exp((19*y)/50))/7500 - (7*exp(-(21*y)/50))/2500 + (7*exp((21*y)/50))/2500 - (23*exp(-(23*y)/50))/7500 + (23*exp((23*y)/50))/7500 - (9*exp(-(27*y)/50))/2500 + (9*exp((27*y)/50))/2500 - (29*exp(-(29*y)/50))/7500 + (29*exp((29*y)/50))/7500 - (31*exp(-(31*y)/50))/7500 + (31*exp((31*y)/50))/7500 - (11*exp(-(33*y)/50))/2500 + (11*exp((33*y)/50))/2500 - (37*exp(-(37*y)/50))/7500 + (37*exp((37*y)/50))/7500 - (13*exp(-(39*y)/50))/2500 + (13*exp((39*y)/50))/2500 - (41*exp(-(41*y)/50))/7500 + (41*exp((41*y)/50))/7500 - (43*exp(-(43*y)/50))/7500 + (43*exp((43*y)/50))/7500 - (47*exp(-(47*y)/50))/7500 + (47*exp((47*y)/50))/7500 - (49*exp(-(49*y)/50))/7500 + (49*exp((49*y)/50))/7500 - exp(-y/100)/7500 + exp(y/100)/7500 - exp(-(3*y)/100)/2500 + exp((3*y)/100)/2500 - (7*exp(-(7*y)/100))/7500 + (7*exp((7*y)/100))/7500 - (3*exp(-(9*y)/100))/2500 + (3*exp((9*y)/100))/2500 - (11*exp(-(11*y)/100))/7500 + (11*exp((11*y)/100))/7500 - (13*exp(-(13*y)/100))/7500 + (13*exp((13*y)/100))/7500 - (17*exp(-(17*y)/100))/7500 + (17*exp((17*y)/100))/7500 - (19*exp(-(19*y)/100))/7500 + (19*exp((19*y)/100))/7500 - (7*exp(-(21*y)/100))/2500 + (7*exp((21*y)/100))/2500 - (23*exp(-(23*y)/100))/7500 + (23*exp((23*y)/100))/7500 - (9*exp(-(27*y)/100))/2500 + (9*exp((27*y)/100))/2500 - (29*exp(-(29*y)/100))/7500 + (29*exp((29*y)/100))/7500 - (31*exp(-(31*y)/100))/7500 + (31*exp((31*y)/100))/7500 - (11*exp(-(33*y)/100))/2500 + (11*exp((33*y)/100))/2500 - (37*exp(-(37*y)/100))/7500 + (37*exp((37*y)/100))/7500 - (13*exp(-(39*y)/100))/2500 + (13*exp((39*y)/100))/2500 - (41*exp(-(41*y)/100))/7500 + (41*exp((41*y)/100))/7500 - (43*exp(-(43*y)/100))/7500 + (43*exp((43*y)/100))/7500 - (47*exp(-(47*y)/100))/7500 + (47*exp((47*y)/100))/7500 - (49*exp(-(49*y)/100))/7500 + (49*exp((49*y)/100))/7500 - (17*exp(-(51*y)/100))/2500 + (17*exp((51*y)/100))/2500 - (53*exp(-(53*y)/100))/7500 + (53*exp((53*y)/100))/7500 - (19*exp(-(57*y)/100))/2500 + (19*exp((57*y)/100))/2500 - (59*exp(-(59*y)/100))/7500 + (59*exp((59*y)/100))/7500 - (61*exp(-(61*y)/100))/7500 + (61*exp((61*y)/100))/7500 - (21*exp(-(63*y)/100))/2500 + (21*exp((63*y)/100))/2500 - (67*exp(-(67*y)/100))/7500 + (67*exp((67*y)/100))/7500 - (23*exp(-(69*y)/100))/2500 + (23*exp((69*y)/100))/2500 - (71*exp(-(71*y)/100))/7500 + (71*exp((71*y)/100))/7500 - (73*exp(-(73*y)/100))/7500 + (73*exp((73*y)/100))/7500 - (77*exp(-(77*y)/100))/7500 + (77*exp((77*y)/100))/7500 - (79*exp(-(79*y)/100))/7500 + (79*exp((79*y)/100))/7500 - (27*exp(-(81*y)/100))/2500 + (27*exp((81*y)/100))/2500 - (83*exp(-(83*y)/100))/7500 + (83*exp((83*y)/100))/7500 - (29*exp(-(87*y)/100))/2500 + (29*exp((87*y)/100))/2500 - (89*exp(-(89*y)/100))/7500 + (89*exp((89*y)/100))/7500 - (91*exp(-(91*y)/100))/7500 + (91*exp((91*y)/100))/7500 - (31*exp(-(93*y)/100))/2500 + (31*exp((93*y)/100))/2500 - (97*exp(-(97*y)/100))/7500 + (97*exp((97*y)/100))/7500 - (33*exp(-(99*y)/100))/2500 + (33*exp((99*y)/100))/2500 + exp(y)/300
So I got a result, but its not the analytical one I wanted, and a bit of a nasty mess. Is it correct?
所以我得到了一个结果，但它不是我想要的分析结果，还有一点糟糕的混乱。是正确的吗?
vpa(subs(res,y,1/2))
ans =
0.34174141693463006644516447861307
I'll copy the analytical result from above so we can compare...
我将从上面复制分析结果以便我们可以比较……
0.34174141687554424792549563431876
As you can see, Simpson's rule, at a step size of 0.01 over [-1,1], did reasonably well, agreeing out to about 9 decimal digits.
正如你所看到的，辛普森的规则，步长为0.01 /(-1,1)，做得相当好，同意约9位小数。
There is no assurance that this technique will work as well on any more general kernel, but it might give you what you desire.
并不能保证这种技术能在任何更通用的内核中发挥作用，但是它可能会给您想要的东西。
#2
1
Looks like you want something like this:
看起来你想要这样的东西:
y = 100; % whatever y is
a = 0;
b = 2;
% you'll need to vectorize the integrand function
f = @(x) x.*x + x.*y
val = quad(f, a, b);
However, if you are looking for an algebraic answer, you'll need to use the Symbolic Toolbox, or some other software, or your calculus book. :-)
但是，如果您正在寻找一个代数的答案，您将需要使用符号工具箱，或者其他一些软件，或者您的微积分书。:-)
The whole "vectorize" thing comes from the Mathworks quad documentation that says:
整个“vectorize”是来自Mathworks quad文档，它说:
The function y = fun(x) should accept a vector argument x and return a vector result y, the integrand evaluated at each element of x.
函数y = fun(x)应该接受一个向量参数x，并返回一个向量结果y，并在x的每个元素上进行积分。