Question

$\int\limits^4_1 \, -3xdx$

Answer 1

Primeiro integro e depois aplico os limites,limite superior menos o inferior : 

Resolvendo ... 

$\int\limits^4_1 {-3x} \, dx \\\\\\- \frac{3x^{1+1}}{1+1} \\\\\\\boxed{- \frac{3x^{2}}{2} }\\\\\\Aplicando\ os\ limites\ :$

$|- \frac{3.4^{2}}{2} |-|- \frac{3.1^{2}}{2} |\\\\\\|- \frac{3.16}{2} |-|- \frac{3.1}{2} |\\\\\\|- \frac{48}{2} |-|- \frac{3}{2}|\\\\- \frac{48}{2} + \frac{3}{2} =\boxed{\boxed{- \frac{45}{2} \ }}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ok$

Answer 2

$\displaystyle\mathsf{\int\limits_{1}^{4}-3x\,dx= \left[-\dfrac{3}{2} {x}^{2} \right] _{1}^{4}} \\\mathsf{-\dfrac{3}{2}. {4}^{2} - \left( - \frac{3}{2}. {1}^{4} \right) =-\dfrac{48}{2} + \dfrac{3}{2} }$

$=\mathsf{-\dfrac{45}{2}}$