Question

integrate limit 2 to 0 1/4+x-x2​

Answer 1

Step-by-step explanation:

$= {\int_0}^{2}dx\: (\frac{1}{4} + x - {x}^{2} ) \\= {\int_0}^{2} \frac{1}{4}dx + {\int_0}^{2}x dx-{\int_0}^{2} {x}^{2} dx \\ = \frac{1}{4} + {[ \frac{ {x}^{2} }{2} ]_0}^{2} - {[ \frac{{x}^{3}}{3} ]_0}^{2} \\ = \frac{1}{4} +[ \frac{ {(2)}^{2} }{2} - \frac{ {(0)}^{2} }{2} ] - [ \frac{ {(2)}^{3} }{3} - \frac{ {(0)}^{3} }{3} ] \\ = \frac{1}{4} + \frac{4}{2} - \frac{8}{3} \\ = \frac{1}{4} + 2 - \frac{8}{3} \\ = \frac{3+12-32}{12} \\ = \frac{ - 17}{12}$

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