Question

Find the area bounded by the curve y = |x − 1|, the x-axis and the lines x = −7 and x =11. (5 marks)

Answer 1

Let's represent the required area on the coordinate plane:

We can see that it consists of two isosceles right triangles with sides 8 and 10, so we can find the required area as a sum of two areas of triangles.

$A=\frac{1}{2}*8*8+\frac{1}{2}*10*10=32+50=82.$

Another way to find the area is using integrals. As our area consists of two parts, we have to split integration area in two areas, separated by point x=1.

$y=|x-1|=\begin{cases} 1-x, & x < 1, \\ x-1, & x\ge 1. \end{cases}$

$A=\int_{-7}^{11} |x-1|dx=\int_{-7}^{1} (1-x)dx+\int_{1}^{11} (x-1)dx=$

$=(x-\frac{x^2}{2}) |_{-7}^{1} +(\frac{x^2}{2}-x) |_{1}^{11} =$

$=(1-\frac{1^2}{2})-(-7-\frac{(-7)^2}{2})+(\frac{11^2}{2}-11)-(\frac{1^2}{2}-1)=$

$=1-\frac{1}{2}+7+\frac{49}{2}+\frac{121}{2}-11-\frac{1}{2}+1=(1+7-11+1)+(-\frac{1}{2}+\frac{49}{2}+\frac{121}{2}-\frac{1}{2})=$

$=-2+\frac{168}{2}=-2+84=82.$

Answer: 82.