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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
The value of the integral $\int\limits^2_{-2}\big|1-\text{x}^2\big|\text{dx}$ is:
  • ✓
    $4$
  • B
    $2$
  • C
    $-2$
  • D
    $0$
✓

Answer

Correct option: A.
$4$
We have,
$\text{I}=\int\limits^2_{-2}\big|1-\text{x}^2\big|\text{dx}$
$\big|1-\text{x}^2\big|=\begin{cases}-\big(1-\text{x}^{2}\big), & -2<\text{x}<-1\\\big(1-\text{x}^2\big), & -1<\text{x}<1 \\-\big(1-\text{x}^2\big),&1<\text{x}<2\end{cases}$
$\therefore\ \text{I}=\int\limits^{-1}_{-2}\big|1-\text{x}^2\big|\text{dx}+\int\limits^1_{-1}\big|1-\text{x}^2\big|\text{dx}+\int\limits^2_1\big|1-\text{x}^2\big|\text{dx}$
$=\int\limits^{-1}_{-2}-\big(1-\text{x}^2)\text{dx}+\int\limits^1_{-1}(1-\text{x}^2)\text{dx}+\int\limits^2_1-(1-\text{x}^2)\text{dx}$
$=-\int\limits^{-1}_{-2}(1-\text{x}^2)\text{dx}+\int\limits^1_{-1}(1-\text{x}^2)\text{dx}-\int\limits^2_1(1-\text{x}^2)\text{dx}$
$=\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^{-1}_{-2}+\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^1_{-1}-\Big[\text{x}-\frac{\text{x}^3}{3}\Big]^2_1$
$=\Big[-1+\frac{1}{3}+2-\frac{8}{3}\Big]+\Big[1-\frac{1}{3}+1-\frac{1}{3}\Big]-\Big[2-\frac{8}{3}-1+\frac{1}{3}\Big]$
$=-\Big[1-\frac{7}{3}\Big]+\Big[2-\frac{2}{3}\Big]-\Big[1-\frac{7}{3}\Big]$
$=-1+\frac{7}{3}+2-\frac{2}{3}-1+\frac{7}{3}$
$=4$

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