Apply the Midpoint Rule to approximate ${\int }_2^{10}(x^2-3x)dx$ using $n=8$ subintervals. Then, calculate the absolute error. The exact value of the integral is $\frac{560}{3}$.

${\int }_2^{10}(x^2-3x)dx\approx 188$

Absolute error: $1.3333$

${\int }_2^{10}(x^2-3x)dx\approx 188$

Absolute error: $0.6667$

${\int }_2^{10}(x^2-3x)dx\approx 186$

Absolute Error: $1.3333$

${\int }_2^{10}(x^2-3x)dx\approx 186$

Absolute error: $0.6667$