The value of - -1 2 4 x 2 - x - dx is equal to

The correct option is A 17 Let I=∫ _ 1 ^ 2 4 x ^ 2 | x | #160; dx =∫ _ 0 ^ 2 4 x ^ 2 ( x) dx+∫ _ 0 ^ 2 4 x ^ 2 (x) dx [ ∵| x | = x ...

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Standard XII

Mathematics

One - One function

The value of ...

Question

The value of $\int_{- 1}^{2} 4 x^{2} | x | d x$ is equal to

17

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Solution

The correct option is A $17$
Let $I = \int_{- 1}^{2} 4 x^{2} | x | d x$
$= \int_{0}^{2} 4 x^{2} (- x) d x + \int_{0}^{2} 4 x^{2} (x) d x$
$[∵ | x | = {\begin{matrix} x, x \geq 0 - x, x < 0 \end{matrix}]$
$= - 4 {[\frac{x^{4}}{4}]}_{- 1}^{0} + 4 {[\frac{x^{4}}{4}]}_{0}^{2}$
$= 1 + 16 = 17$

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The value of ∫2−14x2|x|dx is equal to

The correct option is A 17Let I=∫2−14x2|x|dx=∫204x2(−x)dx+∫204x2(x)dx[∵|x|={x,x≥0−x,x<0]=−4[x44]0−1+4[x44]20=1+16=17

The value of $\int_{- 1}^{2} 4 x^{2} | x | d x$ is equal to

The correct option is A $17$
Let $I = \int_{- 1}^{2} 4 x^{2} | x | d x$
$= \int_{0}^{2} 4 x^{2} (- x) d x + \int_{0}^{2} 4 x^{2} (x) d x$
$[∵ | x | = {\begin{matrix} x, x \geq 0 - x, x < 0 \end{matrix}]$
$= - 4 {[\frac{x^{4}}{4}]}_{- 1}^{0} + 4 {[\frac{x^{4}}{4}]}_{0}^{2}$
$= 1 + 16 = 17$