/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 17

Evaluate the following integrals as they are written. $$\int_{0}^{1} \int_{x}^{1} 6 y d y d x$$

Short Answer

Expert verified
Question: Evaluate the double integral $\iint_{R} 6y \,dx\,dy$, where $R$ is the triangular region with vertices $(0,0)$, $(1,0)$, and $(1,1)$. Answer: The value of the given double integral is 2.

Step by step solution

01

Evaluate the inner integral

The inner integral is given by: $$\int_{x}^{1} 6 y d y$$ To evaluate this integral, we will use basic integration rules: $$\int 6 y d y = 6\int y d y = 6 \frac{y^2}{2} = 3y^2$$ Now, we need to substitute the limits x and 1 to find the inner integral: $$[3y^2]_{x}^{1} = 3(1)^2 - 3(x)^2 = 3 - 3x^2$$
02

Evaluate the outer integral

Now, we will evaluate the outer integral with respect to x: $$\int_{0}^{1} (3 - 3x^2) d x$$ To evaluate this integral, we will again use basic integration rules: $$\int (3 - 3x^2) d x= 3\int d x - 3\int x^2 d x = 3 x - x^3$$ Now, we substitute the limits 0 and 1 to find the outer integral: $$[3 x - x^3]_{0}^{1} = (3 (1) - (1)^3) - (3 (0) - (0)^3) = 3 - 1 = 2$$ So, the value of the given double integral is 2.

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