Home page for accesible maths Math 101 Chapter 5: Integration

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

5.9 Proof of Fundamental Theorem of Calculus

The difference quotient of FF is

F(x+h)-F(x)h=1hxx+hf(t)dt{{F(x+h)-F(x)}\over{h}}={{1}\over{h}}\int_{x}^{x+h}f(t)dt
=1hxx+h(f(t)-f(x))dt+1hxx+hf(x)dt.={{1}\over{h}}\int_{x}^{x+h}(f(t)-f(x))\,dt+{{1}\over{h}}\int_{x}^{x+h}f(x)\,dt.

To prove that F(x)=f(x),F^{\prime}(x)=f(x), we need to prove that the right-hand side converges to f(x)f(x) as h0h\rightarrow 0. Given that ff is continuous, f(t)-f(x)0f(t)-f(x)\rightarrow 0 as h0h\rightarrow 0 and 1hxx+hf(x)dt=f(x){{1}\over{h}}\int_{x}^{x+h}f(x)dt=f(x), so

F(x+h)-F(x)h0+f(x)  (h0),{{F(x+h)-F(x)}\over{h}}\rightarrow 0+f(x)\qquad(h\rightarrow 0),

hence F(x)=f(x)F^{\prime}(x)=f(x).