Economics 1011a: Microeconomic Theory (1a)

Providing an in-depth, explanation of today's course in Macroeconomics.

The fundamental theorem of calculus for differentiable functions allows us in general to compute integrals nicely. You have already made use of this theorem in the homework for today. Earlier in the course, we saw that Sf(x) = h(f(0) + · · · + f(kh)) and Df(x) = (f(x + h) − f(x))/h we have SDf = f(x) − f(0) and DSf(x) = f(x) if x = nh. This now becomes the fundamental theorem. It assumes f 0 to be continuous. R x 0 f 0 (t) dt = f(x) − f(0) and d dx R x 0 f(t) dt = f(x)

If f(x) ≥ 0, then R b a f(x) dx is the area under the graph of f(x) and above the interval [a, b] on the x axes. If the function is negative, then R b a f(x) dx is negative too and the integral is minus the area below the curve: Therefore, R b a f(x) dx is the difference of the area above the graph minus the area below the graph.

We know how to integrate functions like e 6x or 1/(1 + x). The technique of substitution allows to spot more complicated anti-derivatives. If we differentiate the function sin(x 2 ) and use the chain rule, we get cos(x 2 )2x. Indeed, by the fundamental theorem of calculus, the anti-derivative of cos(x 2 )2x is Z cos(x 2 )2x dx = sin(x 2 ) + C .