_-2^2 |x cos ⁡ πx | dx is equal to

Question

Accepted Answer

Correct answer: A. (a) : Let I= _-2^2 |x cos ⁡ πx | dx and f (x )= |x cos ⁡ πx |. Then, f (-x )= |-x cos ⁡ (- πx ) |= |-x cos ⁡ πx | =x cos ⁡ πx=f (x ) So, f (x ) is an even function. ∴I= _-2^2 |x cos ⁡ πx | dx=2 _0^2 |x cos ⁡ πx | dx ( ∵ 0.05em 0.05emf (x ) dx=2 0.05em 0.05emf (x ) dx , iff (-x )=f (x )) Now, f (x )= |x cos ⁡ πx |=(x cos ⁡ πx , if0 x 12-x cos ⁡ πx , if 12<x< 32x cos ⁡ πx , if 32 x 2) ⇒I=2( _0^1/2 0.05emx cos ⁡ πxdx+ _1/2^3/2-x cos ⁡ πxdx+ _3/2^2 0.05emx cos ⁡ πxdx) ⇒I=2(( x sin ⁡ πx+ 1 ^2 cos ⁡ πx)_0^1/2-) (( x sin ⁡ πx+ 1 ^2 cos ⁡ πx)_1/2^3/2+( x sin ⁡ πx+ 1 ^2 cos ⁡ πx)_3/2^2) ⇒I=2 (( 12 +0-(0+ 1 ^2)) ) (-((- 32 +0)-( 12 +0))+((0+ 1 ^2)-(- 32 +0))) ⇒I=2(( 12 - 1 ^2)+( 32 + 12 )+( 1 ^2+ 32 ))= 8

$\int_{- 2}^{2} | x cos πx | dx is equal to$

Solution

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