# Find the area of the region bounded by the lines $y=x-5$ and $x$-axis between the ordinates $x=3$ and $x=7$

Toolbox:
• Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx$ or $\int \limits _a^b y dx$
• If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Step 1:
Area bounded by $y=x-5$ and $x-axis,$ between $x=3,y=7$
Step 2:
Between $x=3,x=5$ the area $A_1$ has on the -ve side of the y-axis and between $x=5,x=7$ the area $A_2$ lies on the +ve side of y-axis
Step 3:
$A=A_1+A_2$
$\quad= \int \limits_3^5 -y dx +\int \limits _5 ^7 y dx$
$\quad=\int \limits_3^5 (5-x) dx +\int \limits_5^7 (x-5) dx$
$\quad= \bigg[5x-\large\frac{x^2}{2}$$\bigg]_3^5 + \bigg [\large\frac{x^2}{2}$$-5x\bigg]_5^7$
$\quad=(25 -\large\frac{25}{2}$$)-(15-\large\frac{9}{2})+\frac{49}{2}$$-35-(\large\frac{25}{2}$$-25)$
$\quad=25-15-35+25-\large\frac{25}{2}+\frac{9}{2}+\frac{49}{2}-\frac{25}{2}$
$\quad=4 \;sq. units$