91Ó°ÊÓ

Here, we are going to evaluate the inner integral first. That is: $$\int_{1}^{4} \frac{3 y}{\sqrt{x+y^{2}}} d x$$ To solve this integral, we can use a substitution method. Let's make a substitution: $$u = x + y^2$$ Then, $$\frac{du}{dx} = 1$$ $$dx = du$$ Now let's find the new limits of integration. When \(x = 1\), \(u = 1 + y^2\). When \(x = 4\), \(u = 4 + y^2\). Now, we can rewrite the integral in terms of u: $$\int_{1+y^2}^{4+y^2} \frac{3 y}{\sqrt{u}} du$$ Now we can integrate with respect to u: $$3y \int_{1+y^2}^{4+y^2} u^{-\frac{1}{2}} du$$ The antiderivative of \(u^{-1/2}\) is \(2u^{1/2}\). So, we have: $$3y \left[ 2u^{\frac{1}{2}} \right]_{1+y^2}^{4+y^2}$$ Now we can substitute back \(u = x + y^2\): $$3y \left[ 2(x+y^2)^{\frac{1}{2}} \right]_{1}^{4}$$ This is the result of the inner integral, now we will plug this result back into the outer integral.

Now we have the following integral to solve: $$\int_{0}^{1} 3y \left[ 2(x+y^2)^{\frac{1}{2}} \right]_{1}^{4} dy$$ We can simplify and plug in the limits of integration for the inner integral: $$\int_{0}^{1} 3y \left[ 2(\sqrt{4+y^2}) - 2(\sqrt{1+y^2}) \right] dy$$ Now, we need to find the antiderivative with respect to y: $$\int_{0}^{1} 6y(\sqrt{4+y^2})dy - \int_{0}^{1} 6y(\sqrt{1+y^2})dy$$ To solve these integrals, we can use substitution method again. For the first integral, let \(v_1 = 4 + y^2\), then \(dv_1 = 2y dy\). For the second integral, let \(v_2 = 1 + y^2\), then \(dv_2 = 2y dy\). Now we have: $$\frac{1}{2}\int_{4}^{5} 6\sqrt{v_1} dv_1 - \frac{1}{2} \int_{1}^2 6\sqrt{v_2} dv_2$$ which results in: $$3 \int_{4}^{5} \sqrt{v_1} dv_1 - 3 \int_{1}^2 \sqrt{v_2} dv_2$$ Now we can solve the remaining integrals: $$3 \left[ \frac{2}{3}(v_1)^{\frac{3}{2}} \right]_{4}^{5} - 3 \left[ \frac{2}{3}(v_2)^{\frac{3}{2}} \right]_{1}^{2}$$ Now plug in the limits of integration and simplify: $$3 \left[ \frac{2}{3}(5)^{\frac{3}{2}} - \frac{2}{3}(4)^{\frac{3}{2}} \right] - 3 \left[ \frac{2}{3}(2)^{\frac{3}{2}} - \frac{2}{3}(1)^{\frac{3}{2}} \right]$$ Now, we obtain the final result: $$2(125-64) - 2(8-1) = 122-14 = 108$$ Thus, the result of the iterated integral is: $$\int_{0}^{1} \int_{1}^{4} \frac{3 y}{\sqrt{x+y^{2}}} d x d y = 108$$