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[tex]\underline{\rm veja~a~figura~que~anexei.}\\\sf a~\acute area~\acute e~dada~por\\\displaystyle\sf A=-\int_0^{\frac{2}{5}}(5x-2)~dx=-\bigg[\dfrac{5}{2}x^2-2x\bigg]_0^{\frac{2}{5}}=\dfrac{5}{2}\cdot\bigg(\dfrac{2}{5}\bigg)^2-2\cdot\dfrac{2}{5}\\\displaystyle\sf\int_0^{\frac{5}{2}}(5x-2)~dx=-\bigg[\dfrac{2\diagup\!\!\!0}{5\diagup\!\!\!0}-\dfrac{4}{5}\bigg]=-\dfrac{-2}{5}\\\huge\boxed{\boxed{\boxed{\boxed{\sf A=\dfrac{2}{5}~u\cdot a}}}}[/tex]
