Use an iterated integral to find the area of the region bounded by $2 x - 5 y = 0 , \quad x + y = 6 , \quad y = 0 .$
A) $\frac { 7 } { 36 }$
B) $\frac { 6 } { 7 }$
C) $\frac { 5 } { 18 }$
D) $\frac { 18 } { 5 }$
E) $\frac { 36 } { 7 }$

Question

Quizplus · Accepted Answer

To find the area of the region bounded by the given lines, we first solve for the intersection points of the lines. The lines are $2x - 5y = 0$, $x + y = 6$, and $y = 0$.1. From $2x - 5y = 0$, we get $y = \frac{2}{5}x$.2. Setting $y = 0$ in $x + y = 6$, we find $x = 6$. This gives us one boundary for the integration.3. To find the intersection of $2x - 5y = 0$ and $x + y = 6$, we substitute $y = \frac{2}{5}x$ into $x + y = 6$ to get $x + \frac{2}{5}x = 6$, which simplifies to $\frac{7}{5}x = 6$, giving $x = \frac{30}{7}$.Now, we set up the integral. The area can be found by integrating with respect to $y$ first (from $y = 0$ to $y = \frac{2}{5}x$) and then with respect to $x$ (from $x = \frac{30}{7}$ to $x = 6$):$$\text{Area} = \int_{\frac{30}{7}}^{6} \int_{0}^{\frac{2}{5}x} dy \, dx$$Solving the inner integral with respect to $y$ gives us:$$\int_{\frac{30}{7}}^{6} \left[\frac{2}{5}x\right] dx = \frac{2}{5} \int_{\frac{30}{7}}^{6} x \, dx$$Integrating $x$ gives us:$$\frac{2}{5} \left[\frac{1}{2}x^2\right]_{\frac{30}{7}}^{6} = \frac{1}{5} \left[x^2\right]_{\frac{30}{7}}^{6} = \frac{1}{5} \left[36 - \left(\frac{30}{7}\right)^2\right]$$Calculating the values gives us:$$\frac{1}{5} \left[36 - \frac{900}{49}\right] = \frac{1}{5} \left[\frac{1764}{49}\right] = \frac{1764}{245} = \frac{36}{7}$$Therefore, the area of the region is $\frac{36}{7}$.

Use an Iterated Integral to Find the Area of the Region