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Chapter 13: Problem 8

Evaluate the following iterated integrals. $$\int_{0}^{3} \int_{-2}^{1}(2 x+3 y) d x d y$$

Short Answer

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Question: Evaluate the iterated integral: $$\int_{0}^{3} \int_{-2}^{1}(2 x+3 y) d x d y$$ Answer: The value of the iterated integral is $$\frac{63}{2}$$.

Step by step solution

01

Evaluate the inside integral with respect to x.

We begin by evaluating the integral inside the brackets: $$\int_{-2}^{1}(2 x+3 y) dx$$ To evaluate this integral, we'll find the antiderivative of the integrand with respect to x and then substitute the limits of integration. The antiderivative of \((2 x+3 y)\) with respect to x is: $$F(x, y) = x^2 + 3xy$$ Now, we'll plug in the limits of integration for x: $$F(1, y) - F(-2, y) = (1^2 + 3(1)y) - ((-2)^2 + 3(-2)y)$$ Simplify this expression to get: $$-3 + 9y$$ Now our expression becomes: $$\int_{0}^{3}(-3+9y) dy$$
02

Evaluate the outside integral with respect to y.

Next, we'll evaluate the remaining integral: $$\int_{0}^{3}(-3+9y) dy$$ To do this, we'll find the antiderivative of the integrand with respect to y, and then substitute the limits of integration. The antiderivative of \((-3+9y)\) with respect to y is: $$G(y) = -3y + \frac{9}{2}y^2$$ Now, we'll plug in the limits of integration for y: $$G(3) - G(0) = (-3(3) + \frac{9}{2}(3)^2) - (0)$$ Simplify this expression, and we get: $$\frac{81}{2} - 9$$ Calculate the result to obtain the final solution: $$\frac{81}{2} - 9 = \frac{63}{2}$$ Therefore, the value of the iterated integral is: $$\int_{0}^{3} \int_{-2}^{1}(2 x+3 y) d x d y = \frac{63}{2}$$

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