Answer to Question #45957 in Integral Calculus for victor

Question

Integrate with respect to x :
∫2−2(x3−3x2+2x−5)dx

Answer 1

∫(2−2(x3−3x2+2x−5))dx=2∫dx-2∫x³dx+2∫3x²dx-2∫2xdx+2∫5dx=2x-2x⁴/4+6x³/3-4x²+10x+C=12x-4x²+2x³ - x⁴/2+C, where C is an arbitrary real constant.