Question

\( \mathop \smallint \nolimits_{ - 2}^2 \left( {sinx + cosx} \right)dx\)
1. 2sin2
2. 2cos2
3. 2
4. 1

Answer 1

Correct Answer - Option 1 : 2sin2

Concept used:

Integral property

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

Calculation:

\( \mathop \smallint \nolimits_{ - 2}^2 \left( {sinx + cosx} \right)dx\)

Let f₁(x) = sinx and f₂(x) = cosx

f₁(x) = sinx

f₁(-x) = sin(-x) = -sinx = -f₁(x)

f₂(x) = cosx

f₂(-x) = cox(-x) = cosx = f₂(x)

By integration property f₁(x) = 0

⇒ \(2\mathop \smallint \nolimits_0^2 cosx\;dx\)

⇒ 2 \(\rm[sinx]^2_0\)

⇒ 2[sin2 - sin0]

⇒ 2sin2