Getting error in bash script; expr $a + 1: integer expression expected

Hi,
I am new to shell/bash script. I am trying to run below script

#!/bin/bash
a=0
b=10
if [ $a == $b ]
then
        echo "a is equal to be"
else
        echo "a is not equal to be"
fi
MAX=10
while [ "$a" -lt $b ]
do
        echo $a
        a='expr $a + 1'
done
 

it is giving below error:

-bash-3.2$ sh cal.sh
a is not equal to be
0
cal.sh: line 11: [: expr $a + 1: integer expression expected
-bash-3.2$

Could you please any one help to solve the problem.
thanks,
Mallik

---------- Post updated at 12:40 PM ---------- Previous update was at 12:12 PM ----------

I have solved above problem by using below code
declare -i a
declare -i b

#!/bin/bash
declare -i a
declare -i b
a=0
b=10
if [ $a == $b ]
then
 echo "a is equal to be"
else
 echo "a is not equal to be"
fi
MAX=10
while [ "$a" -lt $b ]
do
 echo $a
 a=" $a + 1 "
done

---------- Post updated at 12:41 PM ---------- Previous update was at 12:40 PM ----------

solved the above problem

 
declare -i a
declare -i b

Better portable is

a=`expr $a + 1`

(backticks!) or

a=$(($a + 1))

(Posix standard)