`int_0^(2pi) t^2 sin(2t) dt` Evaluate the integral

Wednesday, October 14, 2009

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

You need to
use the integration by parts for $\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$ such that:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

$\displaystyle{u}={{t}}^{{2}}\Rightarrow{d}{u}={2}{t}{\left.{d}{t}\right.}$

$\displaystyle{d}{v}={\sin{{2}}}{t}\Rightarrow{v}=\frac{{-{\cos{{2}}}{t}}}{{2}}$

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}=$
t^2*(-cos 2t)/2|_0^(2pi) + int_0^(2pi) t*cos 2t dt

You need to use the
integration by parts for $\displaystyle{\int_{{0}}^{{{2}\pi}}}{t}\cdot{\cos{{2}}}{t}{\left.{d}{t}\right.}$ such that:

$\displaystyle{u}={t}\Rightarrow{d}{u}$
= dt

$\displaystyle{d}{v}={\cos{{2}}}{t}\Rightarrow{v}=\frac{{{\sin{{2}}}{t}}}{{2}}$

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{t}\cdot{\cos{}}$
2t dt = t*(sin 2t)/2|_0^(2pi) - (1/2)int_0^(2pi) sin 2t dt

$\displaystyle{\int_{{0}}^{{{2}\pi}}}$
t*cos 2t dt = t*(sin 2t)/2|_0^(2pi) + (cos 2t)/4|_0^(2pi)

$\displaystyle{\int_{{0}}^{{{2}\pi}}}$
t^2*sin(2t)dt = t^2*(-cos 2t)/2|_0^(2pi) + t*(sin 2t)/2|_0^(2pi) + (cos 2t)/4|_0^(2pi)

Using the fundamental theorem of calculus yields:

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}={{\left({2}\pi\right)}}^{{2}}\cdot\frac{{-{\cos{{4}}}\pi}}{{2}}+{0}\cdot\frac{{{\cos{{0}}}}}{{2}}+{2}\pi\cdot\frac{{{\sin{{4}}}\pi}}{{2}}-$
0 + (cos 4pi)/4 - (cos 0)/4

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}=-{2}{{\left(\pi\right)}}^{{2}}+\frac{{1}}{{4}}-$
1/4

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}=-{2}{{\left(\pi\right)}}^{{2}}$

Hence, evaluating the integral, using integration by parts,
yields $\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}=-{2}{{\left(\pi\right)}}^{{2}}.$

Spooky Notes

Wednesday, October 14, 2009

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

No comments:

Post a Comment

How is Joe McCarthy related to the play The Crucible?