## Calculus: Early Transcendentals (2nd Edition)

The formula is $$\int_0^axdx=\frac{a^2}{2}.$$
Note that the integrand $x$ is nonnegative on the interval of integration $[0,a]$ so this integral will gives simply the area of the the triangular region bounded by the graph of the function, the $x$ axis and a vertical line $x=a$ (shown on the figure below). The area of this right triangle is just the product of its legs times one half. Since $f(x)=x$ then $f(a)=a$ so the vertical leg has the same length as the horizontal leg which is $a$ so we have $$\int_0^af(x)dx=\frac{1}{2}a\cdot a=\frac{a^2}{2}.$$