@Kitonum

Sir i want to evaluate it for different value for 'h(z)" because in the equation K is very big and i have to change it as k1, k2 so on..

if i will use i=1,2,3  it will be easy .

waiting for your suggestions

for i=1 it should take h1 value and substitute in k[i=1] and evaluate lambda1, similarly 
for i=2 it should take h2 value and substitute in k[i=2] and evaluate lambda2
for i=3 it should take h3 value and substitute in k[i=3] and evaluate lambda3

restart:
h1:=z->1-(delta2/2)*(1 + cos(2*(Pi/L1)*(z - d1 - L1))):
h2:=z->1-(delta2/2)*(1 + cos(2*(Pi/L2)*(z - d2 - L2))):
h3:=z->1+(delta2/2):
K[i]:=((4/h[i](z)^4)-(sin(alpha)/F)-h[i](z)^2+Nb*h[i](z)^4):

lambda1:=Int(K[i],z=0..0.2):
lambda2:=Int(K[i],z=0.2..0.4):
lambda3:=Int(K[i],z=0.4..0.6):
lambda:=lambda1+lambda2+lambda3:
 
F:=0.3:
L1:=0.2:
d1:=0.2:
d2:=0.2:
L2:=0.3:
alpha:=Pi/6:
plot( [seq(eval(lambda, Nb=j), j in [0.1,0.2,0.3])], delta2=0.02..0.1);

@waseem 

my h(z) is defined in different range . based on the range h(z) values substitute in K and integrate.

h:=z->1-(delta2/2)*(1 + cos(2*(Pi/L1)*(z - d1 - L1))): for 0<z<0.2
h:=z->1-(delta2/2)*(1 + cos(2*(Pi/L2)*(z - d2 - L2))): for 0.2<z<0.4
h:=z->1+(delta2/2):                        for 0.4<z<0.6
K:=((4/h(z)^4)-(sin(alpha)/F)-h(z)^2+Nb*h(z)^4):

lambda:= Int(K,z=0..0.2)+ Int(K,z=0.2..0.4)+ Int(K,z=0.4..0.6):

F:=0.3:
L1:=0.2:
d1:=0.2:
d2:=0.2:
L2:=0.3:
alpha:=Pi/6:
plot( [seq(eval(lambda, Nb=j), j in [0.1,0.2,0.3])], delta2=0.02..0.1);

@Kitonum 

sir without writing h1(z), h2(z) and h3(z) is it possible

@tomleslie 

sir  tick mark on all sides please see the attached file.

if we are using axes=box we get only two sides tick marks , i want 4 sides as in the attached figure

@tomleslie sir horizontal range should not be fixed (i,e delta2 values), automatically it should take like the codes i have given above

@tomleslie dear sir  if delta2 values it should take automatically like this i  am attaching the codes in that it will take automatic values if you run the codes  and see the x axis vaues  .

Note please see the x-axis values in the codes given below

restart:
with(DEtools):
with(plots):
eta:= 1+k*x+epsilon*sin(2*Pi*x):
A1:= x-> -exp(-alpha*x)*J^2*R/(2*eta^3+6*xi*J^2*eta^2):
psi0:= A1(x)*y^3:
psi:= delta*psi0:
V:= -diff(psi,x):
delta:= 0.1:
epsilon:= 0.01:
alpha:= 1:
xi:= 0.001:
k:= 0.1:
As:= [0, 2, 4]:
Rs:= [2.0, 5.0, 6.5]:
x:= 0.2:

plots:-display(
     `<|>`(seq(
          plot([seq](
               [eval(V, [J= A, R= r]), y, y= 0..eval(eta, J= A)], A= As),
               title= sprintf("velocity at R=%3.1f", r), labels= ["v", "y"], 
               color= [green, red, blue], 
               linestyle= [solid, dash, dot],
               legend= [seq](J = A, A= As), 
               axes=boxed
          ), r= Rs
     ))
);
restart:
 

@tomleslie thanks for your help, but dont fix delta2 values, then how to draw the graph

@Carl Love 

Thanksss

Dear sir i got the graphs thanks for you help sir i can get a graph of lines style  like this manner i am attaching a file.kindly suggest for the above codes 

@Carl Love 

Dear sir I want a graphs as At R=2; As=0,2,4 one set of graph similarly at R=5 , As=0,2,4 another set of graphs ..... by using do loop or array. 

@Rouben Rostamian  

Dear sir, i want all the graphs one by one at R=2.0 to 5.0 or 6.5  by applying do or for loop or array to execute at the time. 

Dear sir i  want the contour values  automatically display on the contour.

Hear is the code and sample contour plot. 

labelledcontourplot(x^3-y^2,x=-3..3,y=-3..2,axes=boxed);

Dear sir i  want the contour values  automatically display on the contour.

Hear is the code and sample contour plot. 

labelledcontourplot(x^3-y^2,x=-3..3,y=-3..2,axes=boxed);

Dear sir i  want the contour values  automatically display on the contour.

Hear is the code and sample contour plot. 

labelledcontourplot(x^3-y^2,x=-3..3,y=-3..2,axes=boxed);

Dear sir i  want the contour values  automatically display on the contour.

Hear is the code and sample contour plot. 

labelledcontourplot(x^3-y^2,x=-3..3,y=-3..2,axes=boxed);

Thanks a lot for your suggestions