Problem 10 - Calculus Early Transcendentals

Question

Evaluate the following iterated integrals. $$\int_{1}^{3} \int_{1}^{2}\left(y^{2}+y\right) d x d y$$

Step-by-Step Solution

Answer

Question: Evaluate the following iterated integral: $$\int_{1}^{3} \int_{1}^{2}\left(y^{2}+y\right) d x d y$$ Answer: The value of the iterated integral is $$\frac{151}{6}$$.

Step 1: Evaluate the inner integral

First, let's consider the inner integral only, which is with respect to x: $$\int_{1}^{2} (y^2 + y) dx$$ To evaluate this integral, treat y as a constant since we integrate with respect to x. Now integrate each term: $$\left[y^2x + yx\right]_{1}^{2} = (y^2(2) + y(2)) - (y^2(1) + y(1)) = y^2 + y$$

Step 2: Evaluate the outer integral

Now, we need to evaluate the outer integral with respect to y: $$\int_{1}^{3} (y^2 + y) dy$$ Integrate both terms with respect to y: $$\left[\frac{1}{3}y^3 + \frac{1}{2}y^2\right]_{1}^{3} = \left(\frac{1}{3}(3)^3 + \frac{1}{2}(3)^2\right) - \left(\frac{1}{3}(1)^3 + \frac{1}{2}(1)^2\right) = 26 - \frac{5}{6} = \frac{151}{6}$$

Step 3: Final Answer

The value of the iterated integral is: $$\int_{1}^{3} \int_{1}^{2}\left(y^{2}+y\right) d x d y = \frac{151}{6}$$