Evaluate S8.4.3.84 Integral from 1 to 3

[karush]

$\tiny{s8.4.3.84}$
Evaluate $I=\displaystyle\int_1^3 \dfrac{y^3-2y^2-y}{y^2}\ dy$
\begin{array}{lll}\displaystyle
\textit{expand}
&I=\displaystyle\int_1^3 y \ dy
-\displaystyle\int_1^3 2 \ dy
-\displaystyle\int_1^3 \dfrac{1}{y} \ dy
\end{array}

just want to see if I am on the right horse before I cross the stream

 

[skeeter]

interval is $y \in [1,3]$

no division by zero, so it’s all good ... but why not just one definite integral to evaluate

$\displaystyle \int_1^3 y - 2 - \dfrac{1}{y} \, dy$

instead of three?

 

[karush]

interval is $y \in [1,3]$

no division by zero, so it’s all good ... but why not just one definite integral to evaluate

$\displaystyle \int_1^3 y - 2 - \dfrac{1}{y} \, dy$

instead of three?

well don't we have to integrate them individually anyway

integrate
$I=\biggr|\dfrac{y^2}{2}-2y-\ln{y}\biggr|_1^3$
calculate
$I=4-4-\ln (3)=\ln(3)$

this was an even problem number so no book answer

 

[skeeter]

you really need to pay attention to your signs ...

$-\ln(3) = \ln\left(\dfrac{1}{3}\right)$