Evaluate -03 - x-1 x-2 -dx

The correct option is B 11/6 ∫ _ 0 ^ 3 | ( x 1 ) ( x 2 ) nbsp;| dx =∫ _ 0 ^ 1 ( x 1 ) ( x 2 ) dx ∫ _ 1 ^ 2 ( x 1 ) ( x 2 ) dx+∫ _ 2 ^ 3 ( x 1 ...

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Byju's Answer

Standard XII

Mathematics

Multinomial Expansion

Evaluate ∫0...

Question

Evaluate $\int_{0}^{3} | (x - 1) (x - 2) | d x$

\frac{5}{6}

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\frac{11}{6}

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\frac{6}{11}

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\frac{5}{11}

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Solution

The correct option is B $\frac{11}{6}$
$\int_{0}^{3} | (x - 1) (x - 2) | d x = \int_{0}^{1} (x - 1) (x - 2) d x - \int_{1}^{2} (x - 1) (x - 2) d x + \int_{2}^{3} (x - 1) (x - 2) d x = \int_{0}^{1} (x^{2} - 3 x + 2) d x - \int_{1}^{2} (x^{2} - 3 x + 2) d x + \int_{2}^{3} (x^{2} - 3 x + 2) d x = {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x]}_{0}^{1} - {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x]}_{1}^{2} + {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x]}_{2}^{3} = \frac{11}{6}$

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Evaluate ∫30|(x−1)(x−2)|dx

The correct option is B 116∫30|(x−1)(x−2)|dx=∫10(x−1)(x−2)dx−∫21(x−1)(x−2)dx+∫32(x−1)(x−2)dx=∫10(x2−3x+2)dx−∫21(x2−3x+2)dx+∫32(x2−3x+2)dx=[x33−3x22+2x]10−[x33−3x22+2x]21+[x33−3x22+2x]32=116

Evaluate $\int_{0}^{3} | (x - 1) (x - 2) | d x$