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Mathematics Question on Integrals of Some Particular Functions

if  f(x)\ f ( x ) =\bigg \\{ \begin {array} \ e^{\cos x}\sin \ x \\\ 2 \\\ \end {array} \begin {array} \ for |x|\le 2 \\\ \text{otherwise} \\\ \end {array} then ∫−23f(x) dx\int^{3}_{-2} f ( x ) \ dx is equal to

A

0

B

1

C

2

D

3

Answer

2

Explanation

Solution

if  f(x)\ f ( x ) =\bigg \\{ \begin {array} \ e^{\cos x}\sin \ x \\\ 2 \\\ \end {array} \begin {array} \ for |x|\le 2 \\\ \text{otherwise} \\\ \end {array}
∴∫−23 f(x) dx=∫−22f(x) dx+∫23 f(x) dx\therefore \int^{3}_{-2} \ f ( x) \ dx = \int^2_{-2} f ( x ) \ dx + \int^3_2 \ f ( x ) \ dx
                          =∫−22 ecos⁡ x sin⁡ x dx+∫232 dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int^2_{-2} \ e^{\cos \ x } \ \sin \ x \ dx + \int^3_2 2 \ dx
                          =0+2[x]23\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0 + 2 [x]^3_2
                                   [∵ecos⁡ xsin⁡ x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \because e^{\cos \ x }\sin \ x is an odd function ]]
                          =2[3−2]=2                [∵∫−23 f(x) dx=2]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2 [ 3 - 2 ] = 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \because \int^{3}_{-2} \ f ( x ) \ dx = 2 ]