The area bounded by the graphs of f(x) = 4x^3+6x, the x axis and the lines x=1 and x=2 is?
回答 (1)
2015-12-16 1:31 pm
Integrate 4x^3+6x from 1 to 2
Area =
∫ (4x^3+6x) dx = 4 ∫ x^3 dx + 6 ∫ x dx
= (4) (1/4) x^4 + (6) (1/2) x^2
= x^4 + 3x^2
Let F(x) = x^4 + 3x^2
substitute the upper limit 2
F(2) = 2^4 + 3(2^2) = 28
substitute the lower limit 0
F(1) = 1^4 + 3(1^2) = 4
subtract:
F(2)-F(1) = 28-4 = 24
Area =
∫ (4x^3+6x) dx = 4 ∫ x^3 dx + 6 ∫ x dx
= (4) (1/4) x^4 + (6) (1/2) x^2
= x^4 + 3x^2
Let F(x) = x^4 + 3x^2
substitute the upper limit 2
F(2) = 2^4 + 3(2^2) = 28
substitute the lower limit 0
F(1) = 1^4 + 3(1^2) = 4
subtract:
F(2)-F(1) = 28-4 = 24
收錄日期: 2021-04-21 15:50:01
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