I am not sure how to work this out, help would be appreciated thanks :D
The value of b is 1.98
Homework help :D Finding the definite integral?
回答 (5)
2018-07-31 7:47 pm
The integral is |4x^3 - 3x^2| from 1 to 1.98
I = 4(1.98)^3 – 3(1.98)^2 – (4 – 3) = 18.288368 (not negative)
I = 4(1.98)^3 – 3(1.98)^2 – (4 – 3) = 18.288368 (not negative)
2018-07-31 6:17 pm
F(b) = 4 * b^3 - 3 * b^2 ===> If: b = 1.98 ===> F(1.98) = 4 * (1.98)^3 - 3 * (1.98)^2 = 19.288368.
F(1) = 4 * 1^3 - 3 * 1 = 1.
===> F(b) - F(1) = 19.288368 - 1 = 18.288368 (ANSWER).
F(1) = 4 * 1^3 - 3 * 1 = 1.
===> F(b) - F(1) = 19.288368 - 1 = 18.288368 (ANSWER).
2018-07-31 6:14 pm
The integral of 12x²-6x is (12/3)x³-(6/2)x²+C or 4x³-3x²+C using the "power rule" to take the integral.
Evaluate 4x³-3x²+C from 1 to b. That is, if F(x) = 4x³-3x²+C, then what is
F(b)-F(1), or
(4(b)³-3(b)²)-(4(1)³-3(1)²)
Evaluate 4x³-3x²+C from 1 to b. That is, if F(x) = 4x³-3x²+C, then what is
F(b)-F(1), or
(4(b)³-3(b)²)-(4(1)³-3(1)²)
參考: Blessed enough to go to school
2018-07-31 6:11 pm
F(x) = 4x^3 - 3x^2
F(1.98)-F(1) = 4(1.98)^3 - 3(1.98)^2 - (1)=...
F(1.98)-F(1) = 4(1.98)^3 - 3(1.98)^2 - (1)=...
收錄日期: 2021-04-24 01:10:46
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