Notes: `int_0^1 (3t - 1)^50 dt` Evaluate the definite integral.

Tuesday, 8 November 2016

$\displaystyle{\int_{{0}}^{{1}}}{{\left({3}{t}-{1}\right)}}^{{50}}{\left.{d}{t}\right.}$ Evaluate the definite integral.

Given: $\displaystyle{\int_{{0}}^{{1}}}{{\left({3}{t}-{1}\right)}}^{{50}}{\left.{d}{t}\right.}$

Integrate using the Substitution Rule.

Let $\displaystyle{u}={3}{t}-{1}$

$\displaystyle\frac{{{d}{u}}}{{\left.{d}{t}}}={3}$

$\displaystyle{\left.{d}{t}\right.}=\frac{{{d}{u}}}{{3}}$

$\displaystyle={\int_{{0}}^{{1}}}{{u}}^{{50}}\cdot\frac{{{d}{u}}}{{3}}$

$\displaystyle=\frac{{1}}{{3}}{\int_{{0}}^{{1}}}{{u}}^{{50}}{d}{u}$

$\displaystyle=\frac{{1}}{{3}}\cdot\frac{{{u}}^{{51}}}{{51}}$ evaluated from t=0 to t=1

$\displaystyle=\frac{{1}}{{3}}\cdot\frac{{{\left({3}{t}-{1}\right)}}^{{51}}}{{51}}$ evaluated from t=0 to t=1

$\displaystyle=\frac{{1}}{{3}}{\left[\frac{{{\left({3}\cdot{1}-{1}\right)}}^{{51}}}{{51}}-\frac{{{\left({3}\cdot{0}-{1}\right)}}^{{51}}}{{51}}\right]}$

$\displaystyle=\frac{{1}}{{3}}{\left[\frac{{{2}}^{{51}}}{{51}}-\frac{{{{\left(-{1}\right)}}^{{51}}}}{{51}}\right]}$

$\displaystyle=\frac{{1}}{{3}}{\left[\frac{{{{2}}^{{51}}+{{1}}^{{51}}}}{{51}}\right]}$

$\displaystyle=\frac{{{{2}}^{{51}}+{1}}}{{153}}$

$\displaystyle={1.472}\times{{10}}^{{13}}$