<s>[INST] given $ ∫ y dx$ between x=3 and x=1, show that it equals A + B√3 where A and B are integers to be found Where $$∫y dx =2x^{3/2} - 8x^{1/2}$$
I keep getting $6 + 6√3$, which is incorrect. The correct answer should be $6 - 2√3$
But I can't figure out how to correctly derive that answer.
Cheers
 [/INST] It must be a calculation error. Indeed, given the antiderivative expression $2x^{\frac 32} - 8x^{\frac 12}$, there is no need to attempt to find $y$, because the integral between $1$ and $3$ is given by the value of this function at $3$ minus its value at $1$, by the fundamental theorem of calculus.
The value at $3$ :
$$
2 \times 3^{\frac 32}  - 8 \times 3^{\frac 12} = 2 \times 3 \times 3^{\frac 12} - 8 \times 3^{\frac 12} = (6-8) \times 3^{\frac 12}=-2 \times 3^{\frac 12} = -2 \sqrt3
$$
and the value at $1$ :
$$
2 \times 1^{\frac 32} - 8 \times 1 ^{\frac 12} = 2-8 = -6
$$
giving the answer $-2\sqrt 3 -(-6) = 6 - 2 \sqrt 3$.
</s>