find the area enclosed by the graphs of the two curves by integrating with respect to y. y² - 2x = 7 and x…

Answer:

$\frac{125}{6}$

Explanation:

Step1: Rewrite equations for x

From $y^{2}-2x = 7$, we get $x=\frac{y^{2}-7}{2}$. From $x - y=4$, we get $x=y + 4$.

Step2: Find intersection points

Set $\frac{y^{2}-7}{2}=y + 4$. Then $y^{2}-7 = 2y+8$, or $y^{2}-2y - 15=0$. Factoring gives $(y - 5)(y+3)=0$. So $y=-3$ and $y = 5$.

Step3: Set up integral

The area $A=\int_{-3}^{5}[(y + 4)-\frac{y^{2}-7}{2}]dy=\int_{-3}^{5}(y + 4-\frac{y^{2}}{2}+\frac{7}{2})dy=\int_{-3}^{5}(-\frac{y^{2}}{2}+y+\frac{15}{2})dy$.

Step4: Integrate

$\int(-\frac{y^{2}}{2}+y+\frac{15}{2})dy=-\frac{1}{6}y^{3}+\frac{1}{2}y^{2}+\frac{15}{2}y+C$.

Step5: Evaluate definite - integral

$[-\frac{1}{6}(5)^{3}+\frac{1}{2}(5)^{2}+\frac{15}{2}(5)]-[-\frac{1}{6}(-3)^{3}+\frac{1}{2}(-3)^{2}+\frac{15}{2}(-3)]$ $=(-\frac{125}{6}+\frac{25}{2}+\frac{75}{2})-(\frac{9}{2}+\frac{9}{2}-\frac{45}{2})$ $=(-\frac{125}{6}+50)-(-\frac{27}{2})$ $=-\frac{125}{6}+50+\frac{81}{6}$ $=\frac{-125 + 300+81}{6}=\frac{125}{6}$.