$\int_{- 1}^{1} |1 - x| d x =$

$- 2$

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$0$

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$2$

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$4$

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Solution

The correct option is C
$2$

Explanation for correct options:
Finding the value for the given integral:
Consider the given equation as,
$I = \int_{- 1}^{1} |1 - x| d x$
Integrate $I$ as,
$I = \int_{- 1}^{1} (1 - x) d x I = {[x - \frac{x^{2}}{2}]}_{- 1}^{1} I = [(1 - \frac{{(1)}^{2}}{2}) - (- 1 - \frac{{(- 1)}^{2}}{2})] I = (1 - \frac{1}{2} + 1 + \frac{1}{2}) I = 2$
Therefore, the correct answer is Option C.

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The general solution of a differential equation of the type $\frac{d x}{d y} + P_{1} x = Q_{1}$ is

(a) $y e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C$

(b) $y e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C$

(d) $x e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C$

x_{C M} = \frac{\int x d m}{\int d m} = \frac{1 \int 0 x (1 + 2 x) d x}{1 \int 0 (1 + 2 x) d x}

I_{1} = \int_{0}^{1} 2^{x^{2}} d x, I_{2} = \int_{0}^{1} 2^{x^{3}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x

I_{4} = \int_{1}^{2} 2^{x^{3}} d x

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