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Evaluate the following integrals: int_1^2 (4x^3-5x^2+6x+9) dx |
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Answer» Solution :`int_1^2 (4x^3-5x^2+6x+9)DX` =`[4 x^4/4-5 x^3/3 +6 x^2/2 +9x]_1^2` =`[x^4 - (5x^3)/3 +3x^2 +9x]_1^2` =(16-(40)/3 +12 +18)-(1-5/3 +3 +9) =(40-(40)/3)-(13-5/3) = (98)/3 - (34)/3 = (64)/3 |
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