Question #ce8c6

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Question #ce8c6

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Answer 1 · 2017-11-16T16:27:28

$\frac{125}{6} u^{2}$ $125/6u^2$

Explanation:

to find the intersections of both curves, equalize and obtain the cut points

first it will be put in function of y

$x = 5 - y$ $x=5-y$

$x = y^{2} - 4 y + 1$ $x=y^2-4y+1$

equalize

$5 - y = y^{2} - 4 y + 1$ $5-y=y^2-4y+1$

$0 = y^{2} - 3 y - 4$ $0=y^2-3y-4$

factoring

$0 = (y - 4) (y + 1)$ $0=(y-4)(y+1)$

then $y = 4 and y = - 1$ $y=4 and y=-1$

now the area be

$A = \int_{- 1}^{4} (x_{1} - x_{2}) d y$ $A=int_-1^4(x_1-x_2)dy$

where $x_{1} = 5 - y$ $x_1=5-y$ and $x_{2} = y^{2} - 4 y + 1$ $x_2=y^2-4y+1$

$A = \int_{- 1}^{4} (5 - y - y^{2} + 4 y - 1) d y$ $A=int_-1^4(5-y-y^2+4y-1)dy$

$\int_{- 1}^{4} (5) d y = 5 (4) - 5 (- 1) = 25$ $int_-1^4(5)dy=5(4)-5(-1)=25$

$- \int_{- 1}^{4} (y) d y = \frac{y^{2}}{2} \to - (\frac{{(4)}^{2}}{2} - \frac{{(- 1)}^{2}}{2}) = - \frac{15}{2}$ $-int_-1^4(y)dy=y^2/2 -> -((4)^2/2-(-1)^2/2)=-15/2$

$- \int_{- 1}^{4} (y^{2}) d y = \frac{y^{3}}{3} \to - (\frac{{(4)}^{3}}{3} - \frac{{(- 1)}^{3}}{3}) = - \frac{65}{3}$ $-int_-1^4(y^2)dy=y^3/3 -> -((4)^3/3-(-1)^3/3)=-65/3$

$\int_{- 1}^{4} (4 y) d y = 4 (\frac{y^{2}}{2}) \to 4 (\frac{{(4)}^{2}}{2} - \frac{{(- 1)}^{2}}{2})) = 30$ $int_-1^4(4y)dy=4(y^2/2) -> 4((4)^2/2-(-1)^2/2))=30$

$- \int_{- 1}^{4} (- 1) d y = y \to - (4 - (- 1)) = - 5$ $-int_-1^4(-1)dy=y->-(4-(-1))=-5$

$25 - \frac{15}{2} - \frac{65}{3} + 30 - 5 = \frac{125}{6}$ $25-15/2-65/3+30-5=125/6$

$A = \int_{- 1}^{4} (5 - y - y^{2} + 4 y - 1) d y = \frac{125}{6} u^{2}$ $A=int_-1^4(5-y-y^2+4y-1)dy=125/6u^2$

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Question #ce8c6

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