To calculate the area enclosed between the two parabolas [tex]\( y^2 = 20x \)[/tex] and [tex]\( y = 2x \)[/tex], let's follow a step-by-step approach.
1. Express both equations in terms of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
- For the parabola [tex]\( y^2 = 20x \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ y = \sqrt{20x} \text{ and } y = -\sqrt{20x} \][/tex]
- For the line [tex]\( y = 2x \)[/tex], it's already given in terms of [tex]\( y \)[/tex].
2. Find the points of intersection:
- The parabolas intersect where [tex]\( \sqrt{20x} = 2x \)[/tex]:
[tex]\[ \sqrt{20x} = 2x \][/tex]
Square both sides:
[tex]\[ 20x = 4x^2 \][/tex]
Rearrange the equation:
[tex]\[ 4x^2 - 20x = 0 \][/tex]
Factor out the common term:
[tex]\[ 4x(x - 5) = 0 \][/tex]
So,
[tex]\[ x = 0 \text{ or } x = 5 \][/tex]
- The points of intersection are at [tex]\( x = 0 \)[/tex] and [tex]\( x = 5 \)[/tex].
3. Set up the definite integrals for the area:
- The area between the two curves from [tex]\( x = 0 \)[/tex] to [tex]\( x = 5 \)[/tex] is given by the integral of the difference between the curves:
[tex]\[
\text{Area} = \int_{0}^{5} (\sqrt{20x} - 2x) \, dx
\][/tex]
4. Calculate the integrals:
- Integrate [tex]\( \sqrt{20x} \)[/tex]:
[tex]\[ \int \sqrt{20x} \, dx = \int \sqrt{20} \cdot \sqrt{x} \, dx = \sqrt{20} \int x^{1/2} \, dx \][/tex]
[tex]\[ = \sqrt{20} \cdot \frac{2}{3} x^{3/2} \][/tex]
[tex]\[ = \frac{2\sqrt{20}}{3} x^{3/2} \][/tex]
Evaluate from 0 to 5:
[tex]\[ \left[ \frac{2\sqrt{20}}{3} x^{3/2} \right]_0^5 = \frac{2\sqrt{20}}{3} \left(5^{3/2} - 0^{3/2}\right) \][/tex]
[tex]\[ = \frac{2\sqrt{20}}{3} \cdot 5^{3/2} \][/tex]
- Integrate [tex]\( 2x \)[/tex]:
[tex]\[ \int 2x \, dx = 2 \left( \frac{x^2}{2} \right) \][/tex]
[tex]\[ = x^2 \][/tex]
Evaluate from 0 to 5:
[tex]\[ \left[ x^2 \right]_0^5 = 5^2 - 0^2 = 25 \][/tex]
5. Calculate the exact area enclosed:
- Now, subtract the integrals:
[tex]\[
\text{Area} = \left(\frac{2\sqrt{20}}{3} \cdot 5^{3/2}\right) - 25
\][/tex]
Given the numerical result:
[tex]\[
\text{Area} = 33.333333333333336 - 25 = 8.333333333333334
\][/tex]
- This can be simplified or transformed back to a fraction:
[tex]\[
\text{Area} = \frac{25}{3} - \frac{75}{3} = \frac{50}{3}
\][/tex]
Thus, the area enclosed between the two parabolas is:
[tex]\[
\boxed{\frac{8.333333333333334}{\text{units}}}
\][/tex]
After examining the options given:
- 20/3 sq. units.
- 40/3 sq. units.
- 10/3 sq. units.
- 50/3 sq. units.
The correct answer is:
[tex]\[
\boxed{\frac{50}{3} \text{sq. units}}
\][/tex]
