Problem 17 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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 1: 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: $$[3y^2]_{x}^{1} = 3(1)^2 - 3(x)^2 = 3 - 3x^2$$

Step 2: 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.