Finding a definite integral

I want to find the definite integral ,where u and kare both a 1x100 vector and .I want to use the Matlab command "integral" to do it. But to use it, I need the integrand to be a fucntion. The u and k I have are not functions. Is there any way to find this integral? Any help wil be appreciated.

[VBBV]

u and k are both functions of x. But I only know the values of u and k at some points of x.

you can interpolate missing values for u and k by only limited values of x

u = randn(1, 100); % assuming all known values for x
K = randn(1, 100); % assuming all known values for x

X = linspace(0,1,101);
f = @(x) 1.5*exp(-2*(1-x));
for k = 1:length(X)-1    
    F(k) = integral(f,0,X(k+1));
end
S = (u+K)*F.'  % integral sum

S = 1.3599

[Star Strider]

The vectors —

u = [0.1600    0.1600    0.1599    0.1599    0.1597    0.1596    0.1594    0.1592 ...
    0.1590    0.1587    0.1584    0.1581    0.1577    0.1573    0.1569    0.1564 ...
    0.1559    0.1554    0.1548    0.1542    0.1536    0.1529    0.1523    0.1515 ...
    0.1508    0.1500    0.1492    0.1483    0.1475    0.1465    0.1456    0.1446 ...
    0.1436    0.1426    0.1415    0.1404    0.1393    0.1381    0.1369    0.1357 ...
    0.1344    0.1331    0.1318    0.1304    0.1290    0.1276    0.1261    0.1247 ...
    0.1231    0.1216    0.1200    0.1184    0.1167    0.1151    0.1133    0.1116 ...
    0.1098    0.1080    0.1062    0.1043    0.1024    0.1005    0.0985    0.0965 ...
    0.0945    0.0924    0.0903    0.0882    0.0860    0.0838    0.0816    0.0793 ...
    0.0771    0.0747    0.0724    0.0700    0.0676    0.0651    0.0627    0.0601 ...
    0.0576    0.0550    0.0524    0.0498    0.0471    0.0444    0.0417    0.0389 ...
    0.0361    0.0333    0.0304    0.0275    0.0246    0.0216    0.0186    0.0156 ...
    0.0125    0.0095    0.0063    0.0032         0];

k = [0.7543    0.7543    0.7541    0.7537    0.7532    0.7525    0.7517 ...
    0.7507    0.7496    0.7483    0.7469    0.7453    0.7436    0.7417 ...
    0.7397    0.7375    0.7352    0.7328    0.7301    0.7274    0.7245 ...
    0.7214    0.7182    0.7148    0.7113    0.7076    0.7038    0.6998 ...
    0.6957    0.6914    0.6870    0.6824    0.6777    0.6728    0.6677 ...
    0.6626    0.6572    0.6517    0.6461    0.6403    0.6343    0.6282 ...
    0.6219    0.6155    0.6089    0.6022    0.5953    0.5883    0.5811 ...
    0.5737    0.5662    0.5585    0.5507    0.5427    0.5346    0.5263 ...
    0.5178    0.5092    0.5004    0.4915    0.4824    0.4731    0.4637 ...
    0.4541    0.4444    0.4345    0.4244    0.4142    0.4038    0.3933 ...
    0.3825    0.3717    0.3606    0.3494    0.3380    0.3265    0.3148 ...
    0.3029    0.2909    0.2786    0.2663    0.2537    0.2410    0.2281 ...
    0.2150    0.2018    0.1884    0.1748    0.1611    0.1472    0.1331 ...
    0.1188    0.1044    0.0898    0.0750    0.0600    0.0449    0.0295 ...
    0.0140   -0.0016         0];

.

Answer 1

[Torsten]

x = linspace(0,1,101);
u = [0.1600    0.1600    0.1599    0.1599    0.1597    0.1596    0.1594    0.1592 ...
    0.1590    0.1587    0.1584    0.1581    0.1577    0.1573    0.1569    0.1564 ...
    0.1559    0.1554    0.1548    0.1542    0.1536    0.1529    0.1523    0.1515 ...
    0.1508    0.1500    0.1492    0.1483    0.1475    0.1465    0.1456    0.1446 ...
    0.1436    0.1426    0.1415    0.1404    0.1393    0.1381    0.1369    0.1357 ...
    0.1344    0.1331    0.1318    0.1304    0.1290    0.1276    0.1261    0.1247 ...
    0.1231    0.1216    0.1200    0.1184    0.1167    0.1151    0.1133    0.1116 ...
    0.1098    0.1080    0.1062    0.1043    0.1024    0.1005    0.0985    0.0965 ...
    0.0945    0.0924    0.0903    0.0882    0.0860    0.0838    0.0816    0.0793 ...
    0.0771    0.0747    0.0724    0.0700    0.0676    0.0651    0.0627    0.0601 ...
    0.0576    0.0550    0.0524    0.0498    0.0471    0.0444    0.0417    0.0389 ...
    0.0361    0.0333    0.0304    0.0275    0.0246    0.0216    0.0186    0.0156 ...
    0.0125    0.0095    0.0063    0.0032         0];
k = [0.7543    0.7543    0.7541    0.7537    0.7532    0.7525    0.7517 ...
    0.7507    0.7496    0.7483    0.7469    0.7453    0.7436    0.7417 ...
    0.7397    0.7375    0.7352    0.7328    0.7301    0.7274    0.7245 ...
    0.7214    0.7182    0.7148    0.7113    0.7076    0.7038    0.6998 ...
    0.6957    0.6914    0.6870    0.6824    0.6777    0.6728    0.6677 ...
    0.6626    0.6572    0.6517    0.6461    0.6403    0.6343    0.6282 ...
    0.6219    0.6155    0.6089    0.6022    0.5953    0.5883    0.5811 ...
    0.5737    0.5662    0.5585    0.5507    0.5427    0.5346    0.5263 ...
    0.5178    0.5092    0.5004    0.4915    0.4824    0.4731    0.4637 ...
    0.4541    0.4444    0.4345    0.4244    0.4142    0.4038    0.3933 ...
    0.3825    0.3717    0.3606    0.3494    0.3380    0.3265    0.3148 ...
    0.3029    0.2909    0.2786    0.2663    0.2537    0.2410    0.2281 ...
    0.2150    0.2018    0.1884    0.1748    0.1611    0.1472    0.1331 ...
    0.1188    0.1044    0.0898    0.0750    0.0600    0.0449    0.0295 ...
    0.0140   -0.0016         0];
uk = u+k;
fun = @(t)interp1(x,uk,t).*1.5.*exp(-2*(1-t));
I = integral(fun,0,1)

I = 0.2935

I2 = trapz(x,uk.*1.5.*exp(-2*(1-x)))

I2 = 0.2935

[Sharon]

If t and x have the same discretization, can I change t to x, that is,

?

[Torsten]

You don't know in advance what values for t the integrator will choose to evaluate the function. So it's not possible to change t to x.

Another way to evaluate the integral is to use "trapz", but you wrote you want to use "integral".

In case of trapz, the command would be

I = trapz(x,uk.*1.5.*exp(-2*(1-x)))

(see above)

[Sharon]

Thank you very much for your kind reply and providing another method to evaluate the integral. trapz is a good method to do it. I will try it. Which method has a better approximation?

If I use ``integral" and want to separate ``interp1" and the exponential function, can I do the following?

fun1=@(t)interp1(x,uk,t);
fun2=@(t)1.5.*exp(-2*(1-t));
fun3=@(t)fun1(t).*fun2(t);
I=integral(fun3,0,1);

For interp1, do I need to discrete t?

Thank you very much for your help!

[Torsten]

Thank you very much for your kind reply and providing another method to evaluate the integral. trapz is a good method to do it. I will try it. Which method has a better approximation?

If a function is only defined at discrete points, it is reasonable to use an integration method that only needs the values of the function at these discrete points. "trapz" is a method that assumes that the function behaves piecewise linear between the discrete values given. If you supply an interpolation method for the use of "integral" that approximates your discrete values in between with higher accuracy than linear interpolation, you will get a better approximation. But usually the interpolation method is quite arbitrary because you don't know in general how u+k behaves between the discrete values. Thus higher accuracy is hard to achieve in this case (and cannot be verified).

If I use ``integral" and want to separate ``interp1" and the exponential function, can I do the following?

fun1=@(t)interp1(x,uk,t);

fun2=@(t)1.5.*exp(-2*(1-t));

fun3=@(t)fun1(t).*fun2(t);

I=integral(fun3,0,1);

Yes, you can.

For interp1, do I need to discrete t?

If you use "integral", you must provide a function that is defined for all values of x in the interval [0 1]. The easiest way is to use an interpolation function, thus using "interp1". Then instead of discrete values uk(xi) at discrete points xi you now have a function uk(x) that coincides with the array uk at the discrete points, but can be used as input function to "integral".

[Sharon]

Thank you very much for the very detailed explanation! I appreciate it!

Answer 2

[Chunru]

Assuming that u and k are constant vectors:

u = randn(1, 100);
k = randn(1, 100);

syms x
f = (u+k)*int(1.5*exp(-2*(1-x)), [0 1])

f = 

double(f)

ans = 1×100

-0.7007 -0.1411 0.8681 -0.0427 -1.0360 1.5580 -0.8510 -0.7152 -0.3153 -1.1736 0.2003 0.6063 -3.0889 0.2113 0.6936 0.1806 -0.6329 -0.9003 -0.1897 1.5372 0.3675 1.0514 1.3399 0.9247 -0.3171 -0.6689 0.2239 -1.0677 0.4734 -1.1413

[Sharon]

Are you taking (u+k) out?

[Chunru]

Yes, if they are not functions of x.

[Sharon]

Thank you very much for your help and kind reply! u and k are both functions of x. But I only know the values of u and k at some points of x.

[Chunru]

Then you need to define the discrete points x and the corresponding u and k.

[Sharon]

Yes, I have x=linsapce(0,1,101) and u is

u =

Columns 1 through 8

0.1600 0.1600 0.1599 0.1599 0.1597 0.1596 0.1594 0.1592

Columns 9 through 16

0.1590 0.1587 0.1584 0.1581 0.1577 0.1573 0.1569 0.1564

Columns 17 through 24

0.1559 0.1554 0.1548 0.1542 0.1536 0.1529 0.1523 0.1515

Columns 25 through 32

0.1508 0.1500 0.1492 0.1483 0.1475 0.1465 0.1456 0.1446

Columns 33 through 40

0.1436 0.1426 0.1415 0.1404 0.1393 0.1381 0.1369 0.1357

Columns 41 through 48

0.1344 0.1331 0.1318 0.1304 0.1290 0.1276 0.1261 0.1247

Columns 49 through 56

0.1231 0.1216 0.1200 0.1184 0.1167 0.1151 0.1133 0.1116

Columns 57 through 64

0.1098 0.1080 0.1062 0.1043 0.1024 0.1005 0.0985 0.0965

Columns 65 through 72

0.0945 0.0924 0.0903 0.0882 0.0860 0.0838 0.0816 0.0793

Columns 73 through 80

0.0771 0.0747 0.0724 0.0700 0.0676 0.0651 0.0627 0.0601

Columns 81 through 88

0.0576 0.0550 0.0524 0.0498 0.0471 0.0444 0.0417 0.0389

Columns 89 through 96

0.0361 0.0333 0.0304 0.0275 0.0246 0.0216 0.0186 0.0156

Columns 97 through 101

0.0125 0.0095 0.0063 0.0032 0

and k is

k=

Columns 1 through 7

0.7543 0.7543 0.7541 0.7537 0.7532 0.7525 0.7517

Columns 8 through 14

0.7507 0.7496 0.7483 0.7469 0.7453 0.7436 0.7417

Columns 15 through 21

0.7397 0.7375 0.7352 0.7328 0.7301 0.7274 0.7245

Columns 22 through 28

0.7214 0.7182 0.7148 0.7113 0.7076 0.7038 0.6998

Columns 29 through 35

0.6957 0.6914 0.6870 0.6824 0.6777 0.6728 0.6677

Columns 36 through 42

0.6626 0.6572 0.6517 0.6461 0.6403 0.6343 0.6282

Columns 43 through 49

0.6219 0.6155 0.6089 0.6022 0.5953 0.5883 0.5811

Columns 50 through 56

0.5737 0.5662 0.5585 0.5507 0.5427 0.5346 0.5263

Columns 57 through 63

0.5178 0.5092 0.5004 0.4915 0.4824 0.4731 0.4637

Columns 64 through 70

0.4541 0.4444 0.4345 0.4244 0.4142 0.4038 0.3933

Columns 71 through 77

0.3825 0.3717 0.3606 0.3494 0.3380 0.3265 0.3148

Columns 78 through 84

0.3029 0.2909 0.2786 0.2663 0.2537 0.2410 0.2281

Columns 85 through 91

0.2150 0.2018 0.1884 0.1748 0.1611 0.1472 0.1331

Columns 92 through 98

0.1188 0.1044 0.0898 0.0750 0.0600 0.0449 0.0295

Columns 99 through 101

0.0140 -0.0016 0

Answer 3

[Paul]

Hi Sharon,

It seems like we want to compute the following integral

syms x u(x) k(x) w(x)
I = int((u(x)+k(x))*w(x),x,0,1,'Hold',true)

I = 

w(x) has a known form

w(x) = 1.5*exp(-2*(1-x));

which can be substituded into the expression for I

I = subs(I)

I = 

Suppose for now we know the functional form of u(x) and k(x), for example

u(x) = x, k(x) = x^2

u(x) = 

x

k(x) = 

Sub those expressions into I and evaluate the integral

I = subs(I)

I = 

release(I)

ans = 

Now, compute the integral numerically, assuming those same known expressions for u(x) and k(x)

clear
w = @(x) 1.5*exp(-2*(1-x));
u = @(x) x;
k = @(x) x.^2;
integral(@(x) (u(x)+k(x)).*w(x),0,1)

ans = 0.7500

Same result.

But in this question we only have u and k evaluated at discrete set of values of x. For example

xval = linspace(0,1,100);
u = u(xval);
k = k(xval);

Now, to evaluate the integral we need to have some way to approximate u(x) and k(x) based on those vectors of points in order to evaluate the integrand. Here, I'm using interp1 with linear interplolation, but you may want to explore other options after plotting u and k and testing various options, which could be different for u(x) and k(x), to make sure the interpolated values are a a good approximation to the functions u(x) and k(x) (hopefully you have some idea what they should look like). The key is that the interpolation has to be performed in the evaluation of the integrand.

integral(@(x) (interp1(xval,u,x)+interp1(xval,k,x)).*w(x),0,1)

ans = 0.7500