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

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