Page:Calculus Made Easy.pdf/233

Now reckon out the area beneath the curve by counting the little squares below the line, from $x=0$ as far as $x=12$ on the right. There are $18$ whole squares and four triangles, each of which has an area equal to $1{\tfrac {1}{2}}$ squares; or, in total, $24$ squares. Hence $24$ is the numerical value of the integral of ${\dfrac {x}{3}}\,dx$ between the lower limit of $x=0$ and the higher limit of $x=12$.

As a further exercise, show that the value of the same integral between the limits of $x=3$ and $x=15$ is $36$.

(2) Find the area, between limits $x=x_{1}$ and $x=0$, of the curve $y={\dfrac {b}{x+a}}$.

Fig. 55.

${\text{Area}}=\int _{x=0}^{x=x_{1}}y\cdot dx=\int _{x=0}^{x=x_{1}}{\frac {b}{x+a}}\,dx$