Evaluate the integral -22 - xx-1 -dx

The correct option is D 17/3 ∫ _ 2 ^ 2 | x( x 1 ) #160;| #160; dx =∫ _ 2 ^ 0 ( x( x 1 ) #160;) dx ∫ _ 0 ^ 1 ( x( x 1 ) #160;) ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

Evaluate the ...

Question

Evaluate the integral
$\int_{- 2}^{2} | x (x - 1) | d x$

\frac{11}{3}

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\frac{13}{3}

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\frac{16}{3}

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\frac{17}{3}

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Solution

The correct option is D $\frac{17}{3}$
$\int_{- 2}^{2} | x (x - 1) | d x$

$= \int_{- 2}^{0} (x (x - 1)) d x - \int_{0}^{1} (x (x - 1)) d x + \int_{1}^{2} (x (x - 1)) d x$
$= {[\frac{x^{3}}{3} - \frac{x^{2}}{2}]}_{- 2}^{0} - {[\frac{x^{3}}{3} - \frac{x^{2}}{2}]}_{0}^{1} + {[\frac{x^{3}}{3} - \frac{x^{2}}{2}]}_{1}^{2} = \frac{17}{3}$

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Evaluate the integral∫2−2|x(x−1)|dx

The correct option is D 173∫2−2|x(x−1)|dx=∫0−2(x(x−1))dx−∫10(x(x−1))dx+∫21(x(x−1))dx=[x33−x22]0−2−[x33−x22]10+[x33−x22]21=173

Evaluate the integral
$\int_{- 2}^{2} | x (x - 1) | d x$