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Chapter 16: Problem 12

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

Short Answer

Expert verified
Question: Evaluate the double integral of the function \(f(x,y) = y^2+y\) within the given limits: \(1 \le x \le 2\) and \(1 \le y \le 3\). Answer: The value of the iterated integral is 30.

Step by step solution

01

Integrate with respect to x

Integrate the function \(f(x, y)\) with respect to x. In this case f(x, y) = y^2 + y; so we get: $$\int_{1}^{2} (y^{2}+y)~dx$$ Since there are no x terms in the function, the integral with respect to x of y terms will result in giving the x term multiplied by the y terms. So the integral becomes, $$\int_{1}^{2} (y^{2}x+yx)~dx$$ Now, we can evaluate this integral from the limits of x.
02

Evaluate the integral from the limits of x

Now, we evaluate the integral with respect to x using the given limits: $$\left[(y^{2}x+yx) \right]_{1}^{2} = (2^2y^2+2y)-(1^2y^2+y) = (4y^2+2y)-(y^2+y) = 3y^2+y$$ Now, we have obtained a function with respect to y after integrating the given function with respect to x.
03

Integrate with respect to y

Now, we integrate this function with respect to y, using the given limits of y. $$\int_{1}^{3} (3y^2+y)~dy$$ Now, we need to integrate the function \(3y^2+y\) with respect to y. Here is the integral: $$\int(3y^2+y)~dy = y^3+\frac{1}{2}y^2+C$$
04

Evaluate the integral from the limits of y

Now, we evaluate the integral with respect to y using the given limits: $$\left[y^3+\frac{1}{2}y^2 \right]_{1}^{3} = (3^3+\frac{1}{2}(3^2)) - (1^3+\frac{1}{2}(1^2)) = (27+\frac{9}{2}) - (1+\frac{1}{2})$$ Now, we calculate the result: $$\left(27+\frac{9}{2}\right) - \left(1+\frac{1}{2}\right) = 27+4.5-1-0.5 = 30$$ Therefore, the value of the iterated integral is 30.

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