Evaluate the following iterated integrals. int_1^2int_0^1(3x^2+4y^3)dydx

To evaluate the integral: ${\int }_1^2{\int }_0^1(3x^2+4y^3)dydx$
Solution:
When we integrate with respect to one variable then other will be kept as constant.
Evaluating the integral.
${\int }_1^2{\int }_0^1(3x^2+4y^3)dy={\int }_1^2[{\int }_0^1(3x^2+4y^3)dy]dx$
$={\int }_1^2[(3x^{2}y+4\cdot \frac{y^4}{4}{)}_0^1]dx$
$={\int }_1^2[3x^2\cdot 1+1^4-0]dx$
$={\int }_1^2(3x^2+1)dx$
$=[3\cdot \frac{x^3}{3}+x{]}_1^2$
$=[x^3+x{]}_1^2$
$=[(2^3+2)-(1^3+1)]$
$=[12-2]$
$=10$
$\text{Hence, required answer is 10}$