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The value of \(\mathop \smallint \limits_{ - 2}^2 \left\{ {4{x^3} + 2x - 2} \right\}dx\)
1. 8
2. -8
3. 4
4. 4/15
5. None of these

1 Answer

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Best answer
Correct Answer - Option 2 : -8

Concept:

Integral properties:

\(\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}},If\;f\left( { - {\rm{x}}} \right) = f\left( x \right)}\\ {0,\;If\;f\left( { - {\rm{x}}} \right) = - f\left( x \right)} \end{array}} \right.\)

Calculation:

Let \({\rm{I}} = \mathop \smallint \limits_{ - 2}^2 \left\{ {4{x^3} + 2x - 2} \right\}dx\)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_{ - 2}^2 \left( {4{x^3} + 2x} \right)dx - \mathop \smallint \limits_{ - 2}^2 2dx\)

Let I = I1 – I2

Now, \({{\rm{I}}_1} = \mathop \smallint \limits_{ - 2}^2 \left( {4{x^3} + 2x} \right)dx\)

Let’s find if the function is even or odd.

f(x) = 4x3 + 2x

Put x = -x

⇒ f(-x) = 4(-x)3 + 2(-x)

⇒ f(-x) = -(4x3 + 2x) ⇒ f(-x) = - f(x)

Hence, the given function is odd.

∴ I1 = 0

Now, \({{\rm{I}}_2} = \mathop \smallint \limits_{ - 2}^2 2dx\) = \(\left[ {2x} \right]_{ - 2}^2\) = 2 (2 – (-2)) = 2 × 4 = 8

∴ I = I1 – I2 = 0 – 8 = - 8

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