Let A denote the area enclosed by the equations $x = y ^ { 2 }$ and $x = y + 6$ Then A is

A) $\frac { 625 } { 6 }$
B) $\frac { 225 } { 6 }$
C) $\frac { 125 } { 6 }$
D) $\frac { 25 } { 6 }$
E) $\frac { 5 } { 6 }$

Question

Let A denote the area enclosed by the equations $x = y ^ { 2 }$ and $x = y + 6$ Then A is

A) $\frac { 625 } { 6 }$
B) $\frac { 225 } { 6 }$
C) $\frac { 125 } { 6 }$
D) $\frac { 25 } { 6 }$
E) $\frac { 5 } { 6 }$

Quizplus · Accepted Answer

To find the area enclosed by the curves $x = y^2$ and $x = y + 6$, we first find their points of intersection by setting them equal to each other: $y^2 = y + 6$. Solving this quadratic equation gives us $y^2 - y - 6 = 0$, which factors to $(y - 3)(y + 2) = 0$, yielding $y = 3$ and $y = -2$ as the points of intersection. The area A can be found by integrating the difference between the two functions from $y = -2$ to $y = 3$: $A = \int_{-2}^{3} [(y + 6) - y^2] dy$. This simplifies to $\int_{-2}^{3} (6 - y^2 + y) dy$, which when integrated gives $[6y - \frac{y^3}{3} + \frac{y^2}{2}]_{-2}^{3}$. Evaluating this from -2 to 3 gives $18 - \frac{27}{3} + \frac{9}{2} - (-12 + \frac{8}{3} - 2)$, which simplifies to $\frac{125}{6}$.

Let a Denote the Area Enclosed by the Equations x=y2x = y ^ { 2 }x=y2

Let a Denote the Area Enclosed by the Equations $x = y ^ { 2 }$