1.

Evaluate the following integrals: int_1^2 (4x^3-5x^2+6x+9) dx

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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