Given ∫4x^3+4e^2x+k intergrated between the bounds of 3 and 0 equals 2(46+e^6). Find k.
Sorry I couldn't write the question properly in the question box. The question should read:Given ∫304x3+4e2x+k dx = 2(46+e6)Find K.Step 1- Intergrate ∫304x3+4e2x+k x4+2e2x+kx+cStep 2- Sub in bounds (34+2e6+3k+c)-(04+2e0+0k+c)Step 3- Simplify 81+2e6+3k-1Step 4- Equate to Answer 80+2e6+3k = 2(46+e6)Step 5- Simplify k = 4
CM
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