Given, f(x)=4x3−6x2+2x+5,F(0)=5 Anti-derivative of f=F ⇒‌‌F′=f ⇒‌‌F=∫fdx+C =∫(4x3−6x2+2x+5)‌dx+C =[‌
4x4
4
−‌
6x3
3
+‌
2x2
2
+5x]+C ⇒‌‌f=x4−2x3+x2+5x+C On putting x=0, we get, F(0)=5 ⇒‌‌0−2(0)+0+5(0)+C=5 ⇒‌‌C=5 On putting the value of C in Eq. (i), we get F=x4−2x3+x2+5x+5