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The first fundamental theorem of calculus states that (just trying to remember here), with F(x) being the antiderivative of f(x), while u stand for the upper bound and l stands for the lower bound, the definite integral of a function f(x) is equal to F(u) - F(l).
"Solve the integral from 2 to 3 of x^2" Well, this should be easy. Just use the first fundamental theorem of calculus.
The antiderivative of x^2 = (x^3)/3. (3^3)/3 = 9, and (2^3)/3 = 8/3. 9 - 8/3 = 19/3, which is equal to 6.3333333... Time to search this up on good ol' Wolfram Alpha. *Checks* Alright!