Respuesta :
Answer:
18.
Step-by-step explanation:
y^2 = 2x + 6
2x = y^2 - 6
x = 1/2(y^2 - 6)
The points of intersection of the curves are calculated:
y + 1 = 1/2( y^2 - 6)
2y + 2 = y^2 - 6
y^2 - 2y - 8 = 0
( y - 4)(y + 2) = 0
y = 4, -2
when y = 4, x = 5 and when y = -2 x = -1.
Integrating between y = 4 and y = -2
4
INT [ ( y + 1 + 1/2 (6 - y^2) ]dy
-2
= 4
{ y^2 / 2 + y + 3y - y^3/6 }
-2
= 13 1/3 - (-4 2/3)
13 1/3 + 4 2/3
= 18 answer.
The area bounded by the two given curves which are y² = 2x + 6 and x = y + 1 is; 18
How To solve Boundary Integrals?
We want to find the area bounded by the curves;
y² = 2x + 6 and x = y + 1
Put y + 1 for x in first equation to get;
y² = 2(y + 1) + 6
y² - 2y - 8 = 0
Using quadratic equation calculator gives;
y = -2, 4
Now, we can rewrite the first two equations like this;
x = (y²/2) - 3 and x = y + 1
Thus, the area is;
A = [tex]\int\limits^4_{-2}[/tex] [y + 1 - (¹/₂y² - 3)]dy
Integrating this gives us;
A = [4y + ¹/₂y² - ¹/₆y³]⁴₋₂
Solving by plugging in the boundary values gives;
A = 18
Read more about Boundary Integrals at;
https://brainly.com/question/19053586
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