NIMCET 2007 — Mathematics PYQ

The area bounded by a curve $y=f(x)$, the lines $x=a$, $x=b$, and the x-axis is given by the definite integral:

$$\text{Area}={\int }_a^b|f(x)|dx$$

For this problem, the boundaries are from $x=1$ to $x=3$, and the curve is $y=|x-2|$. Since the absolute value function is always non-negative, the bounded area lies entirely above the x-axis.

We can solve this problem using two different methods: Integration or Geometry.

Method 1: Using Integration

Step 1: Definition of the Modulus Function

The absolute value function $y=|x-2|$ changes its behavior at the critical point where $x-2=0⟹x=2$.

• For x < 2: $|x-2|=-(x-2)=2-x$

• For $x\ge 2$: $|x-2|=x-2$

Step 2: Split the Integral

Since our integration limits span from $1$ to $3$, we break the integral at the critical point $x=2$:

$$\text{Area}={\int }_1^3|x-2|dx={\int }_1^2(2-x)dx+{\int }_2^3(x-2)dx$$

Step 3: Compute Each Part

• First Integral:

  $${\int }_1^2(2-x)dx={\left[2x-\frac{x^2}{2}\right]}_1^2$$

  $$=\left(2(2)-\frac{2^2}{2}\right)-\left(2(1)-\frac{1^2}{2}\right)$$

  $$=(4-2)-\left(2-\frac{1}{2}\right)=2-\frac{3}{2}=\frac{1}{2}$$

• Second Integral:

  $${\int }_2^3(x-2)dx={\left[\frac{x^2}{2}-2x\right]}_2^3$$

  $$=\left(\frac{3^2}{2}-2(3)\right)-\left(\frac{2^2}{2}-2(2)\right)$$

  $$=\left(\frac{9}{2}-6\right)-(2-4)=-\frac{3}{2}-(-2)=-\frac{3}{2}+2=\frac{1}{2}$$

Step 4: Add the Areas

$$\text{Total Area}=\frac{1}{2}+\frac{1}{2}=1$$