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# FTC part 1

### Practice

Statement | Justification |
---|---|

Consider the function f(t) on [x,x+h]. There exists at least one u in [x,x+h] where f(u)=m, the absolute minimum of f and at least one v in [x,x+h] where f(v)=M,the absolute maximum of f. | It is given that f is continuous, and so the extreme value theorem can be applied on the closed interval, [x,x+h]. |

So m≤f(t)≤M on [x,x+h] and mh≤the integral from x to x+h of f(t)dt≤Mh | If m≤f(x)≤M for a≤x≤b, then m(b-a)≤the integral from a to b of f(x)dx≤M(b-a). |

So m≤(the integral from x to x+h of f(t)dt)/h≤M So f(u)≤(the integral from x to x+h of f(t)dt)/h≤f(v) So lim h→0 f(u)≤lim h→0 (the integral from x to x+h of f(t)dt)/h≤lim h→0 f(v) So f(x)≤lim h→0 (the integral from x to x+h of f(t)dt)/h≤f(x) | u and v are contained in [x,x+h]. As h approaches 0, u and v must approach x (even "faster"). |

So f(x)=lim h→0 (the integral from x to x+h of f(t)dt)/h. | The squeeze theorem. |

So f(x)=lim h→0 (the integral from a to x+h of f(t)dt)/h - lim h→0 (the integral from a to x of f(t)dt)/h | The integral from a to b of f(x)dx + the integral from b to c of f(x)dx = the integral from a to c of f(x)dx. |

So f(x)=lim h→0 (g(x+h)-g(x))/h and f(x)=g'(x) QED | The definitions of g and its derivative. |

Created by:
Paul C