Question

10 of 10 this quiz: 10 point(s) possible this question: 1 point(s) possible in the figure, the equation of the solid parabola is ( y = x^2 - 6 ) and the equation of the dashed line is ( y = 5x ). determine the area of the shaded region. the area of the shaded region is (square). (type an exact answer.)

Explanation:

Step1: Find Intersection Points

Set \( x^2 - 6 = 5x \), rearrange to \( x^2 - 5x - 6 = 0 \). Factor: \( (x - 6)(x + 1)=0 \), so \( x = 6 \) or \( x=-1 \).

Step2: Set Up Integral

The area between curves is \( \int_{a}^{b} [\text{top curve} - \text{bottom curve}] dx \). Here, top is \( 5x \), bottom is \( x^2 - 6 \), from \( x=-1 \) to \( x = 6 \). So integral is \( \int_{-1}^{6} (5x - (x^2 - 6)) dx=\int_{-1}^{6} (-x^2 + 5x + 6) dx \).

Step3: Integrate

Antiderivative: \( -\frac{x^3}{3}+\frac{5x^2}{2}+6x \).

Step4: Evaluate at Bounds

At \( x = 6 \): \( -\frac{6^3}{3}+\frac{5(6)^2}{2}+6(6)= -72 + 90 + 36 = 54 \). At \( x=-1 \): \( -\frac{(-1)^3}{3}+\frac{5(-1)^2}{2}+6(-1)=\frac{1}{3}+\frac{5}{2}-6=\frac{2 + 15 - 36}{6}=-\frac{19}{6} \). Subtract: \( 54 - (-\frac{19}{6}) = 54+\frac{19}{6}=\frac{324 + 19}{6}=\frac{343}{6} \). Wait, no, wait: Wait, the integral of \( -x^2 +5x +6 \) from -1 to 6: Let's recalculate antiderivative at 6: \( -\frac{216}{3}+\frac{5*36}{2}+36=-72 + 90 + 36 = 54 \). At -1: \( -\frac{(-1)^3}{3}+\frac{5(-1)^2}{2}+6(-1)=\frac{1}{3}+\frac{5}{2}-6=\frac{2 + 15 - 36}{6}=-\frac{19}{6} \). Then \( 54 - (-\frac{19}{6}) = 54+\frac{19}{6}=\frac{324 + 19}{6}=\frac{343}{6} \)? Wait, no, maybe I mixed up top and bottom. Wait, the parabola \( y = x^2 -6 \) and line \( y =5x \). Let's check at x=0: parabola is -6, line is 0. So line is above parabola between x=-1 and x=6? Wait, at x=-1: parabola is \( 1 -6 = -5 \), line is -5. At x=6: parabola is 36 -6=30, line is 30. Wait, so the two curves intersect at x=-1 and x=6, and between them, which is above? Let's take x=0: line is 0, parabola is -6, so line is above. So the integral is \( \int_{-1}^6 (5x - (x^2 -6)) dx = \int_{-1}^6 (-x^2 +5x +6) dx \). Let's compute the integral again. The antiderivative is \( F(x)= -\frac{x^3}{3}+\frac{5x^2}{2}+6x \). Then \( F(6) - F(-1) \). \( F(6)= -\frac{216}{3}+\frac{536}{2}+36= -72 + 90 + 36 = 54 \). \( F(-1)= -\frac{(-1)^3}{3}+\frac{5(-1)^2}{2}+6(-1)= \frac{1}{3}+\frac{5}{2}-6 \). Convert to sixths: \( \frac{2}{6}+\frac{15}{6}-\frac{36}{6}= -\frac{19}{6} \). So \( 54 - (-\frac{19}{6}) = 54 + \frac{19}{6} = \frac{324 + 19}{6} = \frac{343}{6} \)? Wait, that seems off. Wait, maybe I made a mistake in the intersection. Wait, \( x^2 -6 =5x \) => \( x^2 -5x -6=0 \) => (x-6)(x+1)=0, so x=6 and x=-1. Correct. Then, let's check the integrand: 5x - (x² -6) = -x² +5x +6. The integral of -x² is -x³/3, integral of 5x is 5x²/2, integral of 6 is 6x. Correct. Then F(6) = -216/3 + 536/2 + 36 = -72 + 90 + 36 = 54. F(-1) = -(-1)^3/3 + 5(-1)^2/2 + 6(-1) = 1/3 + 5/2 -6. 1/3 is ~0.333, 5/2 is 2.5, sum is 2.833, minus 6 is -3.166, which is -19/6 (~-3.166). So 54 - (-19/6) = 54 + 3.166... = 57.166..., which is 343/6 (~57.166...). Wait, but let's check with another approach. The area between two curves is \( \int_{a}^{b} |f(x) - g(x)| dx \). Since we confirmed f(x)=5x is above g(x)=x² -6 between x=-1 and x=6, the integral is correct. So the calculation seems right.

Answer:

\(\frac{343}{6}\)