Right answer is (d) \(\frac{185}{4}\)
The explanation is: Let \(I=\int_3^45x^3 \,dx\)
F(x)=∫ 5x^3 dx
=\(\frac{5x^4}{4}\)
Applying the limits by using the fundamental theorem of calculus, we get
I=F(4)-F(3)
=\(\frac{5(4)^3}{4}-\frac{5(3)^3}{4}=\frac{5}{4}(4^3-3^3)\)
=\(\frac{5}{4} (64-27)=\frac{5}{4} (37)=\frac{185}{4}\)