To approximate an integral like \int_{a}^{b}f(x)\ dx with Euler's method, you first have to realize, by the Fundamental Theorem of Calculus, that this is the same as calculating F(b)-F(a), where F'(x)=f(x) for all x\in [a,b]. Also note that you can take F(a)=0 and just calculate F(b).
In other words, since Euler's method is a way of approximating solutions of initial-value problems for first-order differential equations, we want to calculate \int_{a}^{b}f(x)\ dx=F(b) by approximating the unique solution of dy/dx=f(x), y(a)=0 at x=b.
This can be done with an "iteration scheme". Pick a positive integer n to be the number of steps of Euler's method you want to use and then let Delta x=(b-a)/n. Once this is done, let x_{0}=a and y_{0}=0 and use the recursive equations x_{k+1}=x_{k}+Delta x, y_{k+1}=y_{k}+Delta y=y_{k}+f(x_{k})\cdot Delta x to generate a sequence of n+1 points (x_{0},y_{0}), (x_{1}, y_{1}), \ldots (x_{n},y_{n}) that approximate the unique solution of the initial-value problem.
The final result is that \int_{a}^{b}f(x)\ dx=F(b)\approx y_{n}.
This can be implemented fairly easily on a calculator or computer, though you'd have to be somewhat experienced with such programming.
As an example, suppose that you want to estimate \int_{0}^{3}x^{2}\ dx (which we already know is 9). The relevant initial-value problem is dy/dx=f(x)=x^2, y(0)=0 and we want to approximate y(3). Let's choose n=4 so that Delta x=\frac{3}{4}=0.75. Then x_{0}=0, x_{1}=0.75, x_{2}=1.5, x_{3}=2.25, and x_{4}=3. Also y_{1}=y_{0}+f(0)\cdot 0.75=0+0=0, y_{2)=y_{1}+f(0.75)\cdot 0.75=0+0.5625\cdot 0.75=0.421875, y_{3}=y_{2}+f(1.5)\cdot 0.75=0.421875+2.25\cdot 0.75=2.109375, and y_{4}=y_{3}+f(2.25)\cdot 0.75=2.109375+3.796875=5.90625 as our approximate answer for the integral.
This would get to be a better approximation (though very slowly) as n increases (and Delta x=(b-a)/n decreases). For instance, if n=100 and Delta x=3/100=0.03, the approximation for the integral is 8.86545.