If R=[0,1]x [0,1], show that 0 ≤ ∫ ∫ R sin ⁡ ( x + y ) d A ≤ 1

Question

If R=[0,1]x [0,1], show that   0  ≤  ∫      ∫          R        sin  ⁡  (  x  +  y  )  d  A  ≤  1

phravincegrln2 · Accepted Answer

If R = [0, 1] x [0, 1], show that   0  ≤  ∫      ∫          R        sin  ⁡  (  x  +  y  )  d  A  ≤  1the region   R  =  (  (  x  ,  y  )      |    0  ≤  x  ≤  1  ,  0  ≤  y  ≤  1  )therefore the integral becomes.      ∫          0              1            ∫          0              1        sin  ⁡  (  x  +  y  )  d  x  d  y  =      ∫          0              1        [  −  cos  ⁡  (  x  +  y  )      ]          0              1        d  y  =  −      ∫          0              1        [  cos  ⁡  (  1  +  y  )  −  cos  ⁡  (  y  )  ]  d  y  =  −  [  sin  ⁡  (  1  +  y  )  −  sin  ⁡  (  y  )      ]          0              1        =  −  [  (  sin  ⁡  2  −  sin  ⁡  1  )  −  (  sin  ⁡  1  −  sin  ⁡  0  )  ]  =  2  sin  ⁡  1  −  sin  ⁡  2  ≅  0.774  0  ≤  0.774  ≤  1

If R=[0,1]x [0,1], show that 0le int int_R sin(x+y)dA le 1

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