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Question
Evaluate the definite integral:
`int_1^2 (4x^3 - 5x^2 + 6x + 9) dx`
Solution
`= [(4x^4)/4 - (5x^3)/3 + (6x^2)/2 + 9x]_1^2`
`= (x^4 - 5/3 x^3 + 3x^2 + 9x)_1^2`
`= (2^4 - 1^4) - 5/3 (2^3 - 1^3) + 3 (2^2 - 1^2) + 9 (2 - 1)`
`= (16 - 1) - 5/3 (8 - 1) + 3 (4 - 1) + 9 (1)`
`= 15 - 35/3 + 9 + 9`
`= 33 - 35/3`
`= (99 - 35)/3`
`= 64/3`
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