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Evaluate the definite integral from 2 to -2: Integral (x^2+3)^2 dx

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svjr | Student, Undergraduate | (Level 2) Honors

Posted November 6, 2011 at 4:12 AM via web

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Evaluate the definite integral from 2 to -2: Integral (x^2+3)^2 dx

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted November 6, 2011 at 5:22 AM (Answer #1)

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`int_-2^2 (x^2 +3)^2 dx = int_-2^2 (x^4 + 6x^2 +9) dx`

`==> int_-2^2 (x^4 + 6x^2 + 9)dx = x^5/5 + 6x^3/3 + 9x`

`==> int_-2^2 (x^4 + 6x^2 + 9) dx = x^5/5 + 2x^3 + 9x`

`==> (-2^5)/5 + 2(-2^3) + 9(-2) = -32/5 + 2*(-8) - 18 = -32/5 -16 -18 = -40.4`

`==> (2^5)/5 + 2(2^3) + 9(2)= 32/5+16+18 = 40.04`

`==>40.4 - (-40.4)= 40.4 + 40.4 = 80.8`

`==> int_-2^2 (x^2+3)^2 dx = 80.8`

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