Chapter 4 - Integration - 4.3 Exercises - Page 276: 72

$$\frac{{{b^3} - {a^3}}}{3}$$

$$\eqalign{ & \int_a^b {{x^2}} dx = \frac{{{b^3} - {a^3}}}{3} \cr & {\text{Using the definition of Definite Integrals }}\left( {{\text{Page 268}}} \right) \cr & \Delta x = \frac{{b - a}}{n}{\text{ and }}{c_i} = a + i\left( {\Delta x} \right),{\text{ }}{c_i} = a + i\left( {\frac{{b - a}}{n}} \right) \cr & \int_a^b {f\left( x \right)} dx = \mathop {\lim }\limits_{\left\| \Delta \right\| \to 0} \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta {x_i} \cr & {\text{Therefore, }} \cr & \int_a^b x dx = \mathop {\lim }\limits_{\left\| \Delta \right\| \to 0} \sum\limits_{i = 1}^n {f\left( {{c_i}} \right)} \Delta {x_i} \cr & {\text{ }} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {{{\left[ {a + i\left( {\frac{{b - a}}{n}} \right)} \right]}^2}} \left( {\frac{{b - a}}{n}} \right) \cr & {\text{ }} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {{a^2} + 2a\left( {\frac{{b - a}}{n}} \right)i + {{\left( {\frac{{b - a}}{n}} \right)}^2}{i^2}} \right]} \left( {\frac{{b - a}}{n}} \right) \cr & {\text{ }} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left[ {{a^2}\left( {\frac{{b - a}}{n}} \right) + 2a{{\left( {\frac{{b - a}}{n}} \right)}^2}i + {{\left( {\frac{{b - a}}{n}} \right)}^3}{i^2}} \right]} \cr & {\text{Properties of a summation}} \cr & {\text{ }} = \mathop {\lim }\limits_{n \to \infty } \left[ {\sum\limits_{i = 1}^n {{a^2}\left( {\frac{{b - a}}{n}} \right)} + \sum\limits_{i = 1}^n {2a{{\left( {\frac{{b - a}}{n}} \right)}^2}} i + \sum\limits_{i = 1}^n {{{\left( {\frac{{b - a}}{n}} \right)}^3}} {i^2}} \right] \cr & {\text{Using the summation Formulas THEOREM 4}}{\text{.2}} \cr & {\text{ }} = \mathop {\lim }\limits_{n \to \infty } \left[ {{a^2}\left( {\frac{{b - a}}{n}} \right)n + 2a{{\left( {\frac{{b - a}}{n}} \right)}^2}\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right)} \right] \cr & {\text{ }} + \mathop {\lim }\limits_{n \to \infty } \left[ {{{\left( {\frac{{b - a}}{n}} \right)}^3}\frac{{n\left( {n + \, 1} \right)\left( {2n + 1} \right)}}{6}} \right] \cr & {\text{ }} = \mathop {\lim }\limits_{n \to \infty } \left[ {{a^2}\left( {b - a} \right) + a{{\left( {b - a} \right)}^2} + \frac{{a{{\left( {b - a} \right)}^2}}}{n}} \right] \cr & + \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{6}{{\left( {b - a} \right)}^3}\left( {\frac{1}{{{n^2}}} + \frac{3}{n} + 2} \right)} \right] \cr & {\text{Evaluate the limit when }}n \to \infty \cr & = {a^2}\left( {b - a} \right) + a{\left( {b - a} \right)^2} + \frac{1}{3}{\left( {b - a} \right)^3} \cr & {\text{Factoring}} \cr & = \left( {b - a} \right)\left[ {{a^2} + a\left( {b - a} \right) + \frac{1}{3}{{\left( {b - a} \right)}^2}} \right] \cr & = \left( {b - a} \right)\left[ {{a^2} + ab - {a^2} + \frac{{{b^2} - 2ab + {a^2}}}{3}} \right] \cr & = \frac{1}{3}\left( {b - a} \right)\left[ {3ab + {b^2} - 2ab + {a^2}} \right] \cr & = \frac{1}{3}\left( {b - a} \right)\left( {{a^2} + ab + {b^2}} \right) \cr & = \frac{{{b^3} - {a^3}}}{3} \cr} $$