Is it possible to find the numerical value of the integral function at a point?

When finding numerical derivatives at a point, you can use the formula $$f'(x)\approx {f(x+h)-f(x)\over{h}}$$

and approximate the derivative with a computer or calculator by plugging into that formula with an extremely small value of $h$. It's also possible to approximate the definite integral with a program by taking the area of increasingly small rectangles. However, is there a way to calculate the value of the integral function at a point?

For example, if you input the formula $f(x)=x^2$, and you input $2$ for the value you want to find, it would give you the value of the function $x^3\over{3}$ at 2, or about 2.6666

Is there a way to do this?

• You use numerical integration, evaluating $\int_0^x t^2 \,dt$, by subdividing the interval $[0,x]$, and using rectangles, or something more efficient, like Simpson's Rule. – André Nicolas Feb 1 '14 at 17:28
• @AndréNicolas Yeah, I realized that while you wrote that comment. Post it as an answer and I'll accept it – scrblnrd3 Feb 1 '14 at 17:30
• You already have answered it, there is no point in my giving the same answer. A related idea is Euler's Method or Runge-Kutta. Take an interval $[a,b]$, let $h=(b-a)/n$, let $y_0=c$ and $y_{n+1}=y_n +h f'(a+(b-a)/n)$. This generates the values of the definite integral for a large number of values $x$ in $[a,b]$. Of course errors accumulate, but with suitable modification you can get useful results. – André Nicolas Feb 1 '14 at 17:43

Oops. I just realized what the value of the value of an integral function at a point would be. Essentially, it would be $$\int_0^bf(x)dx$$