Answer :
To find the area bounded by the curves [tex]\( x = y^2 \)[/tex] and [tex]\( x = -5y - 6 \)[/tex], we need to follow these steps:
1. Identify the Points of Intersection
- First, we need to find the points where the two curves intersect. This means setting the two equations equal to each other and solving for [tex]\( y \)[/tex]:
[tex]\[ y^2 = -5y - 6 \][/tex]
- Rearrange this equation to form a standard quadratic equation:
[tex]\[ y^2 + 5y + 6 = 0 \][/tex]
- Solving this quadratic equation, we find the roots are:
[tex]\[ y = -3 \quad \text{and} \quad y = -2 \][/tex]
- Therefore, the curves intersect at [tex]\( y = -3 \)[/tex] and [tex]\( y = -2 \)[/tex].
2. Set Up the Integral
- The area between the curves can be determined by integrating the difference of the functions [tex]\( x = y^2 \)[/tex] and [tex]\( x = -5y - 6 \)[/tex] with respect to [tex]\( y \)[/tex] over the interval from [tex]\( y = -3 \)[/tex] to [tex]\( y = -2 \)[/tex].
- The integral we need to evaluate is:
[tex]\[ \int_{-3}^{-2} (y^2 - (-5y - 6)) \, dy \][/tex]
- Simplify the integrand:
[tex]\[ \int_{-3}^{-2} (y^2 + 5y + 6) \, dy \][/tex]
3. Evaluate the Integral
- Find the antiderivative of the integrand:
[tex]\[ \int (y^2 + 5y + 6) \, dy = \frac{y^3}{3} + \frac{5y^2}{2} + 6y + C \][/tex]
- Evaluate the antiderivative at the bounds:
[tex]\[ \left[ \frac{-2^3}{3} + \frac{5(-2)^2}{2} + 6(-2) \right] - \left[ \frac{-3^3}{3} + \frac{5(-3)^2}{2} + 6(-3) \right] \][/tex]
4. Substitute and Calculate
- First, calculate at [tex]\( y = -2 \)[/tex]:
[tex]\[ \left[ \frac{(-2)^3}{3} + \frac{5(-2)^2}{2} + 6(-2) \right] = \left[ \frac{-8}{3} + \frac{20}{2} - 12 \right] = \left[ \frac{-8}{3} + 10 - 12 \right] = \left[ \frac{-8}{3} - 2 \right] = \frac{-8}{3} - \frac{6}{3} = \frac{-14}{3} \][/tex]
- Then, calculate at [tex]\( y = -3 \)[/tex]:
[tex]\[ \left[ \frac{(-3)^3}{3} + \frac{5(-3)^2}{2} + 6(-3) \right] = \left[ \frac{-27}{3} + \frac{45}{2} - 18 \right] = \left[ -9 + 22.5 - 18 \right] = \left[ -9 + 4.5 \right] = -4.5 \][/tex]
- Now, subtract these values:
[tex]\[ \frac{-14}{3} - (-4.5) = \frac{-14}{3} + \frac{9}{2} = \frac{-28 + 27}{6} = \frac{-1}{6} \][/tex]
Therefore, the area bounded by the curves [tex]\( x = y^2 \)[/tex] and [tex]\( x = -5y - 6 \)[/tex] is [tex]\( -\frac{1}{6} \)[/tex] or approximately -0.1667 (considering the integration accuracy).
1. Identify the Points of Intersection
- First, we need to find the points where the two curves intersect. This means setting the two equations equal to each other and solving for [tex]\( y \)[/tex]:
[tex]\[ y^2 = -5y - 6 \][/tex]
- Rearrange this equation to form a standard quadratic equation:
[tex]\[ y^2 + 5y + 6 = 0 \][/tex]
- Solving this quadratic equation, we find the roots are:
[tex]\[ y = -3 \quad \text{and} \quad y = -2 \][/tex]
- Therefore, the curves intersect at [tex]\( y = -3 \)[/tex] and [tex]\( y = -2 \)[/tex].
2. Set Up the Integral
- The area between the curves can be determined by integrating the difference of the functions [tex]\( x = y^2 \)[/tex] and [tex]\( x = -5y - 6 \)[/tex] with respect to [tex]\( y \)[/tex] over the interval from [tex]\( y = -3 \)[/tex] to [tex]\( y = -2 \)[/tex].
- The integral we need to evaluate is:
[tex]\[ \int_{-3}^{-2} (y^2 - (-5y - 6)) \, dy \][/tex]
- Simplify the integrand:
[tex]\[ \int_{-3}^{-2} (y^2 + 5y + 6) \, dy \][/tex]
3. Evaluate the Integral
- Find the antiderivative of the integrand:
[tex]\[ \int (y^2 + 5y + 6) \, dy = \frac{y^3}{3} + \frac{5y^2}{2} + 6y + C \][/tex]
- Evaluate the antiderivative at the bounds:
[tex]\[ \left[ \frac{-2^3}{3} + \frac{5(-2)^2}{2} + 6(-2) \right] - \left[ \frac{-3^3}{3} + \frac{5(-3)^2}{2} + 6(-3) \right] \][/tex]
4. Substitute and Calculate
- First, calculate at [tex]\( y = -2 \)[/tex]:
[tex]\[ \left[ \frac{(-2)^3}{3} + \frac{5(-2)^2}{2} + 6(-2) \right] = \left[ \frac{-8}{3} + \frac{20}{2} - 12 \right] = \left[ \frac{-8}{3} + 10 - 12 \right] = \left[ \frac{-8}{3} - 2 \right] = \frac{-8}{3} - \frac{6}{3} = \frac{-14}{3} \][/tex]
- Then, calculate at [tex]\( y = -3 \)[/tex]:
[tex]\[ \left[ \frac{(-3)^3}{3} + \frac{5(-3)^2}{2} + 6(-3) \right] = \left[ \frac{-27}{3} + \frac{45}{2} - 18 \right] = \left[ -9 + 22.5 - 18 \right] = \left[ -9 + 4.5 \right] = -4.5 \][/tex]
- Now, subtract these values:
[tex]\[ \frac{-14}{3} - (-4.5) = \frac{-14}{3} + \frac{9}{2} = \frac{-28 + 27}{6} = \frac{-1}{6} \][/tex]
Therefore, the area bounded by the curves [tex]\( x = y^2 \)[/tex] and [tex]\( x = -5y - 6 \)[/tex] is [tex]\( -\frac{1}{6} \)[/tex] or approximately -0.1667 (considering the integration accuracy).
