Area "=\\int_0^5 (x^2-3x)dx"
"=\\int_0^3(3x-x^2)dx+\\int_3^5(x^2-3x)dx"
"=[3x^2\/2-x^3\/3]_0^3+[x^3\/3-3x^2\/2]_3^5"
"=(27\/2-9)+(5^3\/3-3*5^2\/2)-(9-27\/2)"
"=27-18+(125\/3-75\/2)"
"=(25\/6)+9"
"=79\/6"
Answer: "\\frac{79}{6}."
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