Use the formulas found in your text and the properties of definite integrals to evaluate the following.
Integral (3-top 0-bottom) (7x-2x^3) dx =
any help would be appreciated thx
from 0 to 3 S 7x - 2x^3 dx
(7/2)x^2 - (2/4)x^4 | 0 to 3
[(7/2)(3)^2 - (1/2)(3)^4] - [(7/2)(0)^2 - (1/2)(0)^4]
63/2 - 81/2 - 0
-18/2 => -9
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from 0 to 3 S 7x - 2x^3 dx
(7/2)x^2 - (2/4)x^4 | 0 to 3
[(7/2)(3)^2 - (1/2)(3)^4] - [(7/2)(0)^2 - (1/2)(0)^4]
63/2 - 81/2 - 0
-18/2 => -9