New Notes Blog: `int_1^4(5 - 2t + 3t^2)dt` Evaluate the integral.

$\displaystyle{\int_{{1}}^{{4}}}{\left({5}-{2}{t}+{3}{{t}}^{{2}}\right)}{\left.{d}{t}\right.}$

apply the sum rule and power rule,

$\displaystyle={{\left[{5}{t}-{2}\frac{{{t}}^{{2}}}{{2}}+{3}\frac{{{t}}^{{3}}}{{3}}\right]}_{{1}}^{{4}}}$

$\displaystyle={{\left[{5}{t}-{{t}}^{{2}}+{{t}}^{{3}}\right]}_{{1}}^{{4}}}$

$\displaystyle={\left[{5}\cdot{4}-{{4}}^{{2}}+{{4}}^{{3}}\right]}-{\left[{5}\cdot{1}-{{1}}^{{2}}+{{1}}^{{3}}\right]}$

$\displaystyle={\left({20}-{16}+{64}\right)}-{\left({5}-{1}+{1}\right)}$

$\displaystyle={\left({84}-{16}\right)}-{\left({5}\right)}$

=63

$\displaystyle{\int_{{1}}^{{4}}}{\left({5}-{2}{t}+{3}{{t}}^{{2}}\right)}{\left.{d}{t}\right.}$

apply the sum rule and power rule,

$\displaystyle={{\left[{5}{t}-{2}\frac{{{t}}^{{2}}}{{2}}+{3}\frac{{{t}}^{{3}}}{{3}}\right]}_{{1}}^{{4}}}$

$\displaystyle={{\left[{5}{t}-{{t}}^{{2}}+{{t}}^{{3}}\right]}_{{1}}^{{4}}}$

$\displaystyle={\left[{5}\cdot{4}-{{4}}^{{2}}+{{4}}^{{3}}\right]}-{\left[{5}\cdot{1}-{{1}}^{{2}}+{{1}}^{{3}}\right]}$

$\displaystyle={\left({20}-{16}+{64}\right)}-{\left({5}-{1}+{1}\right)}$

$\displaystyle={\left({84}-{16}\right)}-{\left({5}\right)}$

=63

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