\displaystyle{0} Explanation: Integrate each term using the \displaystyle\text{power rule for integration} \displaystyle\text{Reminder }\ {\left(\overline{{\underline{{{\left|{\left(\frac{{2}}{{2}}\right)}{\left(\int{a}{x}^{{n}}{\left.{d}{x}\right.}=\frac{{a}}{{{n}+{1}}}{x}^{{{n}+{1}}}\right)}{\left(\frac{{2}}{{2}}\right)}\right|}}}}}\right)} ...

Jim H Feb 25, 2017 Here is a limit definition of the definite integral. (I hope it's the one you are using.) . \displaystyle{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}=\lim_{{{n}\rightarrow\infty}}{\sum_{{{i}={1}}}^{{n}}}{f{{\left({x}_{{i}}\right)}}}\Delta{x} ...

Michael B. · Becca M. · Amory W. Sep 2, 2014 When taking integrals, you will normally solve them one term at a time. You will do the inverse of the power rule so the answer would be: \displaystyle{F}{\left({x}\right)}={4}{x}^{{2}}+{3}{x}+{C} ...

If you expand your second answer, you get \frac{(x+1)^2}{2} = \frac{x^2}{2} + x + \frac{1}{2}. This differs from the "correct" answer by a constant. But if two functions differ by a constant, ...

Perform the integration as if the integral were indefinite but without a constant of integration. Subtract the expression evaluated at the lower limit from the expression evaluated at the upper ...

\displaystyle\frac{{1}}{{2}}{x}^{{2}}+{3}{x}+{c} Explanation: Integrate each term using the \displaystyle\text{power rule for integration} \displaystyle\text{Reminder }\ {\left(\overline{{\underline{{{\left|{\left(\frac{{2}}{{2}}\right)}{\left(\int{\left({a}{x}^{{n}}\right)}{\left.{d}{x}\right.}=\frac{{a}}{{{n}+{1}}}{x}^{{{n}+{1}}}\right)}{\left(\frac{{2}}{{2}}\right)}\right|}}}}}\right)} ...