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+4e^2x+k dx = 2(46+e⁶)Find K.Step 1- Intergrate ∫³₀4x³+4e^2x+k x⁴+2e^2x+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