Answer
$\frac{1}{6}a^2$
Work Step by Step
Step 1. Sketch the function as shown in the figure. We can identify the x-intercept at $(a,0)$.
Step 2. The area formed by the function and the x and y axis in the first quadrant is then given by
$A=\int_0^a ydx$
Step 3. Rewrite the function as
$y=(a^{1/2}-x^{1/2})^2=a-2a^{1/2}x^{1/2}+x$
we have
$A=\int_0^a (a-2a^{1/2}x^{1/2}+x)dx=(ax-\frac{4}{3}a^{1/2}x^{3/2}+\frac{1}{2}x^2)|_0^a=a^2-\frac{4}{3}a^{1/2}a^{3/2}+\frac{1}{2}a^2=\frac{1}{6}a^2$
