The area of the region {(x, y): 0 leq x leq f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The region is bounded by the lines $x=3y$,$x+y=2$,$y=0$,and $x=\frac{9}{4}$.First,we find the intersection points of the boundary lines:$1$. Inter

Vedclass · Accepted Answer

$\frac{11}{32}$

Vedclass · Answer

(A) The region is bounded by the lines $x=3y$,$x+y=2$,$y=0$,and $x=\frac{9}{4}$.First,we find the intersection points of the boundary lines:$1$. Intersection of $x=3y$ and $x+y=2$: Substituting $x=3y$ into $x+y=2$ gives $3y+y=2$,so $4y=2$,which means $y=\frac{1}{2}$ and $x=\frac{3}{2}$. Thus,$P = (\frac{3}{2}, \frac{1}{2})$.$2$. Intersection of $x+y=2$ and $y=0$: $x=2$. Thus,$Q = (2, 0)$.$3$. Intersection of $x=\frac{9}{4}$ and $y=0$: $R = (\frac{9}{4}, 0)$.$4$. Intersection of $x=\frac{9}{4}$ and $x=3y$: $y=\frac{x}{3} = \frac{9/4}{3} = \frac{3}{4}$. Thus,$S = (\frac{9}{4}, \frac{3}{4})$.The region is a quadrilateral $PQRS$ with vertices $P(\frac{3}{2}, \frac{1}{2})$,$Q(2, 0)$,$R(\frac{9}{4}, 0)$,and $S(\frac{9}{4}, \frac{3}{4})$.We can calculate the area using integration or by splitting it into simpler shapes. Using the vertical strip method,the area is:Area $= \int_{3/2}^{9/4} (y_{upper} - y_{lower}) dx$Here,$y_{upper}$ is the line $x=3y \implies y=\frac{x}{3}$ for $x \in [3/2, 9/4]$ is incorrect; looking at the region,the upper boundary is $y=x/3$ and the lower boundary is $y=2-x$.Area $= \int_{3/2}^{9/4} (\frac{x}{3} - (2-x)) dx = \int_{3/2}^{9/4} (\frac{4x}{3} - 2) dx$$= [\frac{4x^2}{6} - 2x]_{3/2}^{9/4} = [\frac{2x^2}{3} - 2x]_{3/2}^{9/4}$$= (\frac{2}{3} \cdot \frac{81}{16} - 2 \cdot \frac{9}{4}) - (\frac{2}{3} \cdot \frac{9}{4} - 2 \cdot \frac{3}{2})$$= (\frac{27}{8} - \frac{9}{2}) - (\frac{3}{2} - 3) = (\frac{27-36}{8}) - (\frac{3-6}{2}) = -\frac{9}{8} + \frac{3}{2} = \frac{-9+12}{8} = \frac{3}{8}$.Wait,checking the vertices again: The region is bounded by $x=3y$,$x+y=2$,$y=0$,and $x=9/4$. The area is $\int_{3/2}^{2} (x/3 - (2-x)) dx + \int_{2}^{9/4} (x/3 - 0) dx = \frac{3}{8} + [x^2/6]_2^{9/4} = \frac{3}{8} + (\frac{81}{16 \cdot 6} - \frac{4}{6}) = \frac{3}{8} + (\frac{27}{32} - \frac{2}{3}) = \frac{3}{8} + \frac{81-64}{96} = \frac{36+17}{96} = \frac{53}{96}$.Re-evaluating the integral: The region is bounded by $x=3y$ (upper),$x+y=2$ (lower left),$y=0$ (lower right),$x=9/4$ (right). The area is $\int_{3/2}^{9/4} (x/3) dx - \int_{3/2}^{2} (2-x) dx = [x^2/6]_{3/2}^{9/4} - [2x - x^2/2]_{3/2}^{2} = (\frac{81}{16 \cdot 6} - \frac{9}{4 \cdot 6}) - ((4-2) - (3 - 9/8)) = (\frac{27}{32} - \frac{3}{8}) - (2 - 15/8) = \frac{15}{32} - \frac{1}{8} = \frac{15-4}{32} = \frac{11}{32}$.

The area of the region $\{(x, y): 0 \leq x \leq \frac{9}{4}, 0 \leq y \leq 1, x \geq 3y, x+y \geq 2\}$ is

Similar Questions

The area bounded by the curve $x^2 = 8y$ and the straight line $x - 8y + 2 = 0$ is

The area bounded by the parabola $y^{2}=x$,the straight line $y=4$,and the $y$-axis in square units is:

The area of the curve $xy^2 = a^2(a - x)$ bounded by the $y$-axis is

If $x^{2}+y^{2}=a^{2}$,then $\int_{0}^{a} \sqrt{1+\left(\frac{dy}{dx}\right)^{2}} dx=$

The value of $a$ $(a > 0)$ for which the area bounded by the curves $y = \frac{x}{6} + \frac{1}{x^2}$,$y = 0$,$x = a$ and $x = 2a$ has the least value,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module