\input epsf
\input graphicx
\magnification 2000
\vsize 10.7truein
\hsize 7.5truein
\hoffset -0.75truein
\voffset -0.7truein
\nopagenumbers
\parskip 10pt
\parindent 0 pt

Use the Riemann sum definition of integral to evaluate $\int_2^5 x^3 dx$

$\Delta x = {5 - 2 \over n} = {3 \over n} $, $x_i = 2 + {3i \over n} $, $f(x) = x^3$.

$\int_2^5 x^3 dx = lim_{n \rightarrow \infty} \Sigma_{i = 1}^n f(x_i) \Delta x$
$ = lim_{n \rightarrow \infty} \Sigma_{i = 1}^n (2 + {3i \over n})^3 ({3 \over n} ) $

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

$ lim_{n \rightarrow \infty} \Sigma_{i = 1}^n [8 +  {36i \over n} + {54i^2 \over n^2} + {27i^3 \over n^3}] ({3 \over n} )$

$ lim_{n \rightarrow \infty} \Sigma_{i = 1}^n [{24 \over n} +  {108i \over n^2} + {162i^2 \over n^3} + {81i^3 \over n^4}] $

$ lim_{n \rightarrow \infty} [ \Sigma_{i = 1}^n   {24 \over n} +   \Sigma_{i = 1}^n   {108i \over n^2} + \Sigma_{i = 1}^n    {162i^2 \over n^3} +  \Sigma_{i = 1}^n   {81i^3 \over n^4}] $

$ lim_{n \rightarrow \infty} [ {24 \over n}   \Sigma_{i = 1}^n   1 +   {108 \over n^2} \Sigma_{i = 1}^n   {i}   +  {162 \over n^3} \Sigma_{i = 1}^n    {i^2} +   {81 \over n^4}  \Sigma_{i = 1}^n   {i^3}  ] $

$ lim_{n \rightarrow \infty} [ {24}  +   {108 \over n^2}  ({n(n+1) \over 2}) 
  +  {162 \over n^3} ({n(n+1)(2n+1) \over 6}) +   {81 \over n^4}  ({n(n+1) \over 2})^2  ] $

$ lim_{n \rightarrow \infty} [ {24}  +   {108 \over n}  ({n+1 \over 2}) 
  +  {162 \over n^2} ({2n^2+3n+1 \over 6}) +   {81 \over n^2}  ({n^2 + 2n + 1 \over 4})  ] $

$ lim_{n \rightarrow \infty} [ {24}  +   {108 \over n}  ({n(1 + {1 \over n}) \over 2}) 
  +  {162 \over n^2} ({n^2(2+{3 \over n}+{1 \over n^2}) \over 6}) +   {81 \over n^2}  ({n^2(1  + {2 \over n} + {1 \over n^2}) \over 4})  ] $

$ lim_{n \rightarrow \infty} [ {24}  +   {54}  {(1 + {1 \over n}) }
  +  {27 } ({(2+{3 \over n}+{1 \over n^2}) }) +   {81}  ({(1  + {2 \over n} + {1 \over n^2}) \over 4})  ] $

$=  [ {24}  +   {54}  {(1) }
  +  {27 } ({2}) +   {81}  ({1  \over 4})  ] $
$=  [{ {24}(4)  +   {54}  {(4) }
  +  {27} ({8 }) +   {81}   \over 4}  ] $

$=  { 96  +   216 + 216 +   81  \over 4} $
$=  { 609 \over 4} $

 Check:  $\int_2^5 x^3 dx = {x^4 \over 4}|_2^5 = {5^4 - 2^4 \over 4} = {625 - 16 \over 4} = { 609 \over 4}$


$\Sigma_{i = 1}^n   {i}   = {n(n+1) \over 2} $, \hfil
$\Sigma_{i = 1}^n   {i}^2   = {n(n+1)(2n+1) \over 6} $, \hfil 
$\Sigma_{i = 1}^n   {i}^3   = [{n(n+1) \over 2}]^2 $


\end
5.5 Examples.

1.)  $\int 2x e^{x^2} dx$

2.)  $\int 3x^2 \sqrt{x^3 + 1} dx$

3.)  $\int {x dx \over x^2 + 4}$

4.)  $\int x \sqrt{1 + x} dx$

5.)  $\int \sqrt{x}(x^2 - 1)dx$

6.)  $\int cos^3(x) dx$


HW 13

[5.2) 21, 23, 24 (if you didn't turn in with HW12)]

3.10) 20, 22, 26, 32


5.4)  23, 45, 47, 50, 54

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

and...
\end

$(a + b)^3 = (a^2 + 2ab + b^2)(a + b) = a^3 + 3a^2b + 3ab^2 + b^3$

$a^3 + 2a^2b + ab^2 
          a^2b + 2ab^2 + b^3$

\end