First Fundamental Theorem of Calculus

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The First Fundamental Theorem of Calculus

The First Fundamental Theorem of Calculus: Statement and Proof

The First Fundamental Theorem of Calculus

Let f be a continuous function on [a,b] and let

F'(x) = f(x)

then

$ \int_a^b f(x) dx = F(b) - F(a) $

Proof:

Cut up the interval [a,b] into several pieces with

a = x₀ < x₁ < x₂ < x₃ < ... < x_n-1 < x_n = b

Then

F(b) - F(a) =

[F(x_n) - F(x_n-1)] + [F(x_n-1) - F(x_n-2)] + [F(x_n-2) - F(x_n-3)] +... + [F(x₂) - F(x₁)] + [F(x₁) - F(x₀)]

= $ \sum_{i=1}^{n} [F(x_i) - F(x_{i-1})] $

By the mean value theorem there is a c_i between x_i-1 and x_i with

                         F(x_i) - F(x_i-1)               F(x_i) - F(x_i-1)
        F'(c_i) =                                  =
                           x_i   -   x_i-1 $\Delta$x_i

Multiplying both sides by $\Delta$x_i gives

F'(c_i)$\Delta$x_i = F(x_i) - F(x_i-1)

Substituting into the sum gives

$ \sum_{i=1}^{n} F'(c_i) \Delta x_i $

Taking the limit as n approaches infinity, gives the definite integral.

Examples

Example 1:

$ \int_{1}^{2} 2x dx = x^2 |_1^2 = 4 - 1 = 3 $

Example 2:

Find the area bounded by the curve

y = x² - x , y = 0, x = 4

$Graph of x^2 - 2x. Between x = 0 and 2 it I below the x-axis. Past x=2 it is above the x-axis.$

Notice that there is area both below the x-axis and above. We can find:

$\int_{i=0}^{4} (x^2 - 2x) dx = \int_{i=2}^{4} (x^2 - 2x) dx - \int_{i=0}^{2} (x^2 - 2x) dx $

$ (\frac{x^3}{3} - x^2)|_2^4 - (\frac{x^3}{3} - x^2)|_0^2 $

$ = [ (\frac{64}{3} - 16) - (\frac{8}{3} - 4) ] \ - [(\frac{8}{3} - 4) - (0)] $

$ = 8$

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\( \int_a^b f(x) dx = F(b) - F(a) \)