## University Calculus: Early Transcendentals (3rd Edition)

$\frac{1}{2}$
Because $1\lt2$, by Equation (2): $\int_{1}^{\sqrt 2} x dx=\frac{(\sqrt 2)^2}{2}-\frac{1^2}{2}=\frac{2}{2}-\frac{1}{2}=\frac{1}{2}$