Problem 45 - Calculus Early Transcendentals

Question

Evaluate the following iterated integrals. $$\int_{0}^{1} \int_{1}^{4} \frac{3 y}{\sqrt{x+y^{2}}} d x d y$$

Step-by-Step Solution

Answer

The value of the given double integral is $2\sqrt{125} - 4\sqrt{16} - 2\sqrt{8} + 2$.

Step 1: Integrate with respect to x

To begin, we will integrate the given function with respect to x, keeping in mind the limits of integration for x. $$\int_{1}^{4} \frac{3 y}{\sqrt{x+y^{2}}} d x$$ To integrate this function, we perform a substitution. Let \(u = x + y^2\). Then, \(du = dx\). Now, we need to update the limits of integration. When x = 1, the new variable u = 1 + y^2, and when x = 4, u = 4 + y^2. Our new integral is: $$\int_{1 + y^2}^{4 + y^2} \frac{3 y}{\sqrt{u}} d u$$ Now, integrate with respect to u. $$3y \int_{1 + y^2}^{4 + y^2} u^{-1/2} d u$$ Using the power rule for integration, we get: $$3y\left[2u^{1/2}\right]_{1+y^2}^{4+y^2}$$ Now, substitute back the x terms: $$3y\left[2\sqrt{x + y^2}\right]_{x=1}^{x=4}$$ Now evaluate the integral at the limits of integration: $$3y(2\sqrt{4+y^2} - 2\sqrt{1+y^2})$$ Our new function is: $$F(y) = 3y(2\sqrt{4+y^2} - 2\sqrt{1+y^2})$$

Step 2:checked

Now, we will integrate this function with respect to y and use the limits of integration for y. $$\int_{0}^{1} 3y(2\sqrt{4+y^2} - 2\sqrt{1+y^2}) dy$$ Since we have a difference of two functions, we will split the integrals: $$2\int_{0}^{1} 3y\sqrt{4+y^2}\, dy - 2\int_{0}^{1} 3y\sqrt{1+y^2}\, dy$$ At this point, we'll need to perform another substitution for both integrals. First integral: Let \(v = 4 + y^2\). Then, \(dv = 2y\, dy\). Second integral: Let \(w = 1 + y^2\). Then, \(dw = 2y\, dy\). Updating the limits for the integrals: First integral: When y = 0, \(v = 4\) and when y = 1, \(v = 5\). Second integral: When y = 0, \(w = 1\) and when y = 1, \(w = 2\). Our new integrals are: $$2\int_{4}^{5} \frac{3}{2}\sqrt{v}\, dv - 2\int_{1}^{2} \frac{3}{2}\sqrt{w}\, dw$$ Now integrate both integrals: $$3\left[\frac{2}{3}v^{3/2}\right]_{4}^{5} - 3\left[\frac{2}{3}w^{3/2}\right]_{1}^{2}$$ Now evaluate the integrals at their respective limits: $$3\left(\frac{2}{3}(5^{3/2} - 4^{3/2})\right) -3\left(\frac{2}{3}(2^{3/2} - 1^{3/2})\right)$$ Simplify the expression: $$2(5^{3/2} - 4^{3/2}) - 2(2^{3/2} - 1^{3/2})$$ Now compute the final value: $$2(\sqrt{125} - 2\sqrt{16}) - 2(\sqrt{8} - 1)$$ $$2\sqrt{125} - 4\sqrt{16} - 2\sqrt{8} + 2$$ Thus, the solution for the given iterated integral is: $$\int_{0}^{1} \int_{1}^{4} \frac{3 y}{\sqrt{x+y^{2}}} d x d y = 2\sqrt{125} - 4\sqrt{16} - 2\sqrt{8} + 2$$