Question

The area of the region enclosed by the parabolas \( y = x^2 - 5x \) and \( y = 7x - x^2 \) is _________.

Answer 1

Correct Answer: 72

Given curves:

\( y = x^2 - 5x \quad \text{and} \quad y = 7x - x^2 \)

Let

\( f(x) = x^2 - 5x \quad \text{and} \quad g(x) = 7x - x^2 \)

To find the area enclosed between these curves, we calculate:

\( \int_0^6 (g(x) - f(x)) \, dx = \int_0^6 ((7x - x^2) - (x^2 - 5x)) \, dx \)

Simplify the integrand:

\( = \int_0^6 (12x - 2x^2) \, dx \)

Now, integrate term by term:

\( = \left[ \frac{12x^2}{2} - \frac{2x^3}{3} \right]_0^6 \)

Substitute the limits:

\( = (6 \cdot 6^2) - \frac{2}{3} \cdot (6)^3 \)

\( = 216 - 144 = 72 \, \text{unit}^2 \)

Answer 2

Step 1: Understand the curves and their intersection
We are given two parabolas: \[ y_1=x^2-5x\quad(\text{opens upward}),\qquad y_2=7x-x^2\quad(\text{opens downward}). \] The enclosed region lies between their points of intersection. Solve \(x^2-5x=7x-x^2\): \[ 2x^2-12x=0\;\Longrightarrow\;2x(x-6)=0\;\Longrightarrow\;x=0,\;6. \] Thus, the region is between \(x=0\) and \(x=6\).

Step 2: Decide which curve is on top
Pick a test point, say \(x=3\): \[ y_1(3)=3^2-5\cdot3=9-15=-6,\qquad y_2(3)=7\cdot3-3^2=21-9=12. \] Hence \(y_2\) is above \(y_1\) on \([0,6]\).

Step 3: Set up the area integral
Area \[ A=\int_{0}^{6}\big(y_2-y_1\big)\,dx =\int_{0}^{6}\Big[(7x-x^2)-(x^2-5x)\Big]\,dx =\int_{0}^{6}(12x-2x^2)\,dx. \]

Step 4: Integrate and evaluate
\[ A=\left[6x^2-\frac{2}{3}x^3\right]_{0}^{6} =\left(6\cdot 36-\frac{2}{3}\cdot 216\right)-0 =216-144 =72. \]

Step 5: Quick sanity check
The integrand \(12x-2x^2=2x(6-x)\) is nonnegative on \([0,6]\) and symmetric about \(x=3\), consistent with a positive, moderate area. The computed value fits the geometry in the sketch.

Final answer

72