Notes: `int_0^1 root(3)(1 + 7x) dx` Evaluate the definite integral.

Sunday, 26 February 2017

Given $\displaystyle{\int_{{0}}^{{1}}}{\sqrt[{{3}}]{{{1}+{7}{x}}}}{\left.{d}{x}\right.}$

Integrate using the u-substitution method.

Let $\displaystyle{u}={1}+{7}{x}$

$\displaystyle\frac{{{d}{u}}}{{\left.{d}{x}}}={7}$

$\displaystyle{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{7}}$

$\displaystyle={\int_{{0}}^{{1}}}{{u}}^{{\frac{{1}}{{3}}}}\cdot\frac{{{d}{u}}}{{7}}$

$\displaystyle=\frac{{1}}{{7}}{\int_{{0}}^{{1}}}{{u}}^{{\frac{{1}}{{3}}}}{d}{u}$

$\displaystyle=\frac{{1}}{{7}}\cdot\frac{{{u}}^{{\frac{{4}}{{3}}}}}{{\frac{{4}}{{3}}}}$ Evaluated from x=0 to x=1.

$\displaystyle=\frac{{1}}{{7}}\cdot\frac{{3}}{{4}}\cdot{{\left({1}+{7}{x}\right)}}^{{\frac{{4}}{{3}}}}$ Evaluated from x=0 to x=1.

$\displaystyle=\frac{{3}}{{28}}$
[(1+7*1)^(4/3)-(1+7*0)^(4/3)]

$\displaystyle=\frac{{3}}{{28}}{\left[{{8}}^{{\frac{{4}}{{3}}}}-{{1}}^{{\frac{{4}}{{3}}}}\right]}$

$\displaystyle=\frac{{3}}{{28}}{\left[{16}-{1}\right]}$

$\displaystyle=\frac{{3}}{{28}}{\left[{15}\right]}$

$\displaystyle=\frac{{45}}{{28}}$

$\displaystyle={1.607}$

Notes