Evaluate a definite integral of a power function using FTC Part 2

First rewrite the integrand using negative exponents: $\frac{2}{x^3} = 2x^{-3}$. Applying the power rule for antiderivatives, $\int 2x^{-3}\,dx = 2\cdot \frac{x^{-2}}{-2} = -x^{-2} = -\frac{1}{x^2}$. So $F(x) = -\frac{1}{x^2}$ is an antiderivative of $2/x^3$.

By FTC Part 2, $\int_1^2 \frac{2}{x^3}\,dx = F(2) - F(1) = \left(-\frac{1}{2^2}\right) - \left(-\frac{1}{1^2}\right) = -\frac{1}{4} + 1 = \frac{3}{4}$. Thus the exact value of the integral is $\frac{3}{4}$.

The function used was F of x equals negative one over x squared, the antiderivative of two over x cubed found by the power rule. By FTC Part 2, this antiderivative was evaluated at the upper limit two and the lower limit one, then subtracted. This gave negative one fourth minus negative one, yielding three fourths.