To find the definite integral of the function (48x^3 - 6x + 11) from -1 to 0, we first need to find the antiderivative of the function.
The antiderivative of (48x^3 - 6x + 11) is:
[16x^4 - 3x^2 + 11x + C]
To evaluate the definite integral from -1 to 0, we substitute these values into the antiderivative:
[16(0)^4 - 3(0)^2 + 11(0) - (16(-1)^4 - 3(-1)^2 + 11(-1))]
Simplifying further:
[0 + 0 + 0 - (16 - 3 - 11)][= 0 - (-8)][= 8]
Therefore, the definite integral of the function (48x^3 - 6x + 11) from -1 to 0 is 8.
To find the definite integral of the function (48x^3 - 6x + 11) from -1 to 0, we first need to find the antiderivative of the function.
The antiderivative of (48x^3 - 6x + 11) is:
[16x^4 - 3x^2 + 11x + C]
To evaluate the definite integral from -1 to 0, we substitute these values into the antiderivative:
[16(0)^4 - 3(0)^2 + 11(0) - (16(-1)^4 - 3(-1)^2 + 11(-1))]
Simplifying further:
[0 + 0 + 0 - (16 - 3 - 11)]
[= 0 - (-8)]
[= 8]
Therefore, the definite integral of the function (48x^3 - 6x + 11) from -1 to 0 is 8.