Area under y = k·√x

Text

Compute the area under the curve $y=3\sqrt{x}$ between $x=0$ and $x=4$, and check that it is $\tfrac{2}{3}$ of the bounding rectangle.

Solution

$A=\int_{0}^{4}3x^{1/2}\,dx=3\cdot\frac{2}{3}\left[x^{3/2}\right]_{0}^{4}=2\cdot 8=16.$ The rectangle of base $4$ and height $y(4)=6$ has area $24$, and $\dfrac{16}{24}=\dfrac{2}{3}$, as expected. $\boxed{A=16=\tfrac{2}{3}\cdot 24}$