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$\int_2^6 (-{1 \over 2}t + 4)dt$ = 
$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n f(t_i ) \Delta t$

$\Delta t = {6 - 2 \over n} = {4 \over n}$

$t_i = 2 + i \Delta t = 2 + {4i \over n}$ (using right-hand endpoints)

{$\int_2^6 (-{1 \over 2}t + 4)dt$ $= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n f(2 + {4i \over n} ) ({4 \over n}) $}

$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n [ -{1 \over 2}(2 + {4i \over n}) + 4 ] ({4 \over n})$

$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n [ -1  - {2i \over n} + 4 ] ({4 \over n})$

$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n [ 3  - {2i \over n} ] ({4 \over n})$


$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n [ {12 \over n}  - {8i \over n^2} ]$

$= lim_{n \rightarrow \infty} (\Sigma_{i = 1}^n {12 \over n}  - \Sigma_{i = 1}^n {8i \over n^2})$

$= lim_{n \rightarrow \infty} (12  -{8\over n^2}\Sigma_{i = 1}^n i)$

$= lim_{n \rightarrow \infty} (12  - {8\over n^2}  {n(n+1) \over 2})$

$= lim_{n \rightarrow \infty} (12  -  {4n^2+4n \over n^2})$

$= lim_{n \rightarrow \infty} (12  -  4 - {4 \over n})$
  = 8

\vfill
Show them Gauss's derivation:

~~ 1~  +  ~2~~~  + ... + ~n-1~ + n

~~ n~  + n-1~ + ... +  ~~2~~~  + 1

-------------------------------------------------------

n+1 + n+1 + ... + n+1 + n+1= n(n+1)

Thus $\Sigma_{i = 1}^n i = {n(n+1) \over 2}$


\eject

5.4) Indefinite integral:

$\int f(x) dx =$ anti-derivative of $f$.

I.e, $\int f(x) dx = F(x)$ where $F'(x) = f(x)$

Ex: 

1.)  $\int x^3 dx =$  

 
2.)  $\int [3x^4 + 2 + {3 \over x} + cos(x)] dx =$  

5.5)  Integration by substitution

If $F(x) = ln|x^2 - x|$, then by chain rule $F'(x) = {1 \over x^2 - x}(2x - 1) = {2x - 1 \over x^2 - x} $


Thus $\int  {2x - 1 \over x^2 - x} dx = $

How do you recognize derivative when chain rule involved?  u -substition

$\int  {2x - 1 \over x^2 - x}dx  = \int {du \over u} = \int {1 \over u}du = ln|u| + C =ln| x^2 - x| + C  $

Let $u = x^2 - x$, then $du = 2x - 1$

\end

$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n f(a + {(b - a)i \over n} ) ({b-a \over n}) $

$= lim_{n \rightarrow \infty} \Sigma_{i = 1}^n f(a + {(b - a)i \over n} ) ({b-a \over n}) $

HW11

5.2) 18, 21, 23, 24, 30, 34, 35, 38, 39, 49, 50, 53

5.5) 8, 9, 11, 13, 16, 21, 28, 31, 36, 41, 42

and

5.3) 3, 7-10, 13-16,19, 37, 38, 61, 62



HW12

 5.5)  56, 57, 61, 64, 66, 76




HW 13:
5.4) 23, 45, 47, 50, 54


6.1) 9, 19

HW 14:
6.1) 24, 28, 41
6.2) 31-36 all
6.3) 22, 26

Recommended
6.3) 21-26 all 

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