NIMCET 2007 — Mathematics PYQ

To solve a definite integral containing an absolute value function, we must first determine where the expression inside the modulus changes its sign.

Step 1: Analyze the Modulus Function

The expression inside the absolute value is $1-x^2$. Let's find its critical points by setting it to zero:

$$1-x^2=0⟹x^2=1⟹x=\pm 1$$

Now, we check the sign of $1-x^2$ in the intervals between our integration limits $[-2,3]$:

• For $x\in [-2,-1)$: 1 - x^2 < 0 \implies |1 - x^2| = -(1 - x^2) = x^2 - 1

• For $x\in [-1,1]$: $1-x^2\ge 0⟹|1-x^2|=1-x^2$

• For $x\in (1,3]$: 1 - x^2 < 0 \implies |1 - x^2| = -(1 - x^2) = x^2 - 1

Step 2: Split the Integral

Using the intervals determined above, we split the original integral into three parts:

$${\int }_{-2}^3|1-x^2|dx={\int }_{-2}^{-1}(x^2-1)dx+{\int }_{-1}^1(1-x^2)dx+{\int }_1^3(x^2-1)dx$$

Step 3: Integrate Each Section

Part 1: ${\int }_{-2}^{-1}(x^2-1)dx$

$${\left[\frac{x^3}{3}-x\right]}_{-2}^{-1}=\left(\frac{(-1{)}^3}{3}-(-1)\right)-\left(\frac{(-2{)}^3}{3}-(-2)\right)$$

$$=\left(-\frac{1}{3}+1\right)-\left(-\frac{8}{3}+2\right)$$

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

Part 2: ${\int }_{-1}^1(1-x^2)dx$

Since $1-x^2$ is an even function, we can simplify this to $2{\int }_0^1(1-x^2)dx$:

$$2{\left[x-\frac{x^3}{3}\right]}_0^1=2\left(1-\frac{1}{3}\right)=2\left(\frac{2}{3}\right)=\frac{4}{3}$$

Part 3: ${\int }_1^3(x^2-1)dx$

$${\left[\frac{x^3}{3}-x\right]}_1^3=\left(\frac{3^3}{3}-3\right)-\left(\frac{1^3}{3}-1\right)$$

$$=(9-3)-\left(\frac{1}{3}-1\right)$$

$$=6-\left(-\frac{2}{3}\right)=6+\frac{2}{3}=\frac{20}{3}$$

Step 4: Sum the Values

Now, add the results of the three parts together:

$$\text{Total Value}=\frac{4}{3}+\frac{4}{3}+\frac{20}{3}=\frac{4+4+20}{3}=\frac{28}{3}$$

Final Answer

The correct option is (c) 28/3.