Determine the area of the shaded region in the figure.
x
y
y equals 4 xy=4xy equals x squared minus 5y=x2−5

A coordinate system has an unlabeled horizontal x-axis and an unlabeled vertical y-axis. A curve labeled y equals x squared minus 5 is a parabola that opens upward from its vertex on the negative y-axis. A line labeled y equals 4 x rises from left to right and passes through the origin. The curve and line intersect at two points, one in the third quadrant and one in the first quadrant. The region below the x-axis, above the parabola, and above the line in the third quadrant is shaded.
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Part 1
The area of the shaded region is enter your response here.
​(Type an exact​ answer.)

To determine the area of the shaded region in the given figure, we need to find the points of intersection between the line \( y = 4x \) and the parabola \( y = x^2 - 5 \). Then we integrate the difference of the functions over the interval defined by these points.

First, set the equations equal to each other to find the points of intersection:
\[ 4x = x^2 - 5 \]

Rearrange the equation:
\[ x^2 - 4x - 5 = 0 \]

Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} \]
\[ x = \frac{4 \pm \sqrt{16 + 20}}{2} \]
\[ x = \frac{4 \pm \sqrt{36}}{2} \]
\[ x = \frac{4 \pm 6}{2} \]

Thus, we have two solutions:
\[ x = \frac{4 + 6}{2} = 5 \]
\[ x = \frac{4 - 6}{2} = -1 \]

The points of intersection are \( x = -1 \) and \( x = 5 \).

Next, determine the y-values at these points of intersection:
\[ y = 4(-1) = -4 \]
\[ y = 4(5) = 20 \]

Therefore, the points of intersection are \( (-1, -4) \) and \( (5, 20) \).

To find the area of the shaded region in the third quadrant (between \( x = -1 \) and \( x = 0 \)), we set up the integral of the difference between the parabola and the line:
\[ \text{Area} = \int_{-1}^{0} [(x^2 - 5) - 4x] \, dx \]
\[ = \int_{-1}^{0} (x^2 - 4x - 5) \, dx \]

Evaluate the integral:
\[ \int_{-1}^{0} (x^2 - 4x - 5) \, dx = \left[ \frac{x^3}{3} - 2x^2 - 5x \right]_{-1}^{0} \]

Calculate the definite integral:
\[ \left[ \frac{(0)^3}{3} - 2(0)^2 - 5(0) \right] - \left[ \frac{(-1)^3}{3} - 2(-1)^2 - 5(-1) \right] \]
\[ = [0] - \left[ -\frac{1}{3} - 2 + 5 \right] \]
\[ = 0 - \left( -\frac{1}{3} - 2 + 5 \right) \]
\[ = 0 - \left( -\frac{1}{3} + 3 \right) \]
\[ = 0 + \frac{1}{3} - 3 \]
\[ = \frac{1}{3} - 3 \]
\[ = \frac{1 - 9}{3} \]
\[ = -\frac{8}{3} \]

The area of the shaded region is \( \boxed{\frac{8}{3}} \).

Do you want more details or have any questions?

Here are some related questions you might find helpful:
1. What is the significance of finding the points of intersection in such problems?
2. How do you solve quadratic equations using the quadratic formula?
3. Can you explain the process of setting up an integral to find the area between curves?
4. How do you evaluate a definite integral?
5. What is the geometrical interpretation of the area under a curve?
6. How do you determine the bounds of integration for a region in the coordinate plane?
7. What is the importance of subtracting the functions when finding areas between curves?
8. How can this method be applied to find the area between other types of curves?

**Tip:** Always double-check the limits of integration and the functions involved when setting up integrals to ensure accuracy in finding the area.

Calculate the area of the shaded region between a parabola and a line in the coordinate plane. Learn how to find points of intersection, set up integrals, and evaluate definite integrals step-by-step. Suitable for advanced high school students.

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