91Ó°ÊÓ

To perform the inner integral, we'll evaluate: $$\int_{0}^{1} x^{5} y^{2} e^{x^{3} y^{3}} d y$$ Let's use integration by substitution: Let \(u = x^{3} y^{3}\). Then \(\frac{du}{dy} = 3x^{3} y^{2}\), or \(dy = \frac{du}{3x^{3} y^{2}}\). Substitute u and dy into the integral: $$\int x^{5} y^{2} e^{u} \frac{du}{3x^{3} y^{2}}$$ Simplify the expression: $$\frac{1}{3} x^{2} \int e^{u} du$$ Now integrate with respect to u: $$\frac{1}{3} x^{2}\left[ e^{u} \right]_{0}^{x^{3}}$$ Substitute y back in for u: $$\frac{1}{3} x^{2}\left[ e^{x^{3} y^{3}} \right]_{0}^{1}$$ Evaluate the expression at the limits of integration: $$\frac{1}{3} x^{2}\left( e^{x^{3}} - e^{0} \right) = \frac{1}{3} x^{2}\left( e^{x^{3}} - 1 \right)$$ Now we have the inner integral evaluated, and can move on to the outer integral.

To evaluate the outer integral: $$\int_{0}^{2} \frac{1}{3} x^{2} \left( e^{x^{3}} - 1 \right) dx$$ Separate the terms inside the integral: $$\frac{1}{3} \int_{0}^{2} x^{2} e^{x^{3}} dx - \frac{1}{3} \int_{0}^{2} x^{2} dx$$ For the first integral, we can use integration by substitution again: Let \(v = x^3\), then \(\frac{dv}{dx} = 3x^2\), or \(dx = \frac{dv}{3x^2}\). Substitute v and dx into the integral: $$\frac{1}{3} \int e^{v} \frac{dv}{3x^2}$$ Simplify the expression: $$\frac{1}{9} \int e^{v} dv$$ Now integrate with respect to v: $$\frac{1}{9}\left[ e^{v} \right]_{0}^{2}$$ Substitute x back in for v: $$\frac{1}{9}\left[ e^{x^{3}} \right]_{0}^{2}$$ Evaluate the expression at the limits of integration: $$\frac{1}{9}\left( e^{8} - e^{0} \right) = \frac{1}{9}\left( e^{8} - 1 \right)$$ For the second integral, integrate the \(x^2\) term directly: $$-\frac{1}{3} \int_{0}^{2} x^{2} dx = -\frac{1}{3}\left[ \frac{1}{3}x^{3} \right]_{0}^{2} = -\frac{8}{9}$$ Now we combine the results of the two integrals: $$\frac{1}{9}(e^{8}-1) - \frac{8}{9}$$ So, the evaluated iterated integral is: $$\int_{0}^{2} \int_{0}^{1} x^{5} y^{2} e^{x^{3} y^{3}} d y d x = \frac{1}{9}(e^{8}-1) - \frac{8}{9}$$