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Write down the definite integral that represents the enclosed area AND find the area. f(x)=-(x-3)(x+2), the vertical axis and the horizontal axis

Write down the definite integral that represents the enclosed area AND find the area. f(x)=−(x−3)(x+2) f(x)=-(x-3)(x+2) , the vertical axis and the horizontal axis

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Full solution

Q. Write down the definite integral that represents the enclosed area AND find the area. f(x)=−(x−3)(x+2) f(x)=-(x-3)(x+2) , the vertical axis and the horizontal axis
  1. Find Intersecting Points: To find the area enclosed by the curve and the axes, we first need to determine the points where the curve intersects the horizontal axis. These points are the roots of the equation f(x)=−(x−3)(x+2)=0f(x) = -(x-3)(x+2) = 0.
  2. Set Up Definite Integral: Setting f(x)f(x) to zero gives us the equation 0=−(x−3)(x+2)0 = -(x-3)(x+2). We can find the roots by setting each factor equal to zero: x−3=0x-3 = 0 and x+2=0x+2 = 0.
  3. Calculate Integral: Solving these equations gives us the roots x=3x = 3 and x=−2x = -2. These are the points where the curve intersects the xx-axis.
  4. Evaluate Upper Limit: The area enclosed by the curve and the axes will be between these two xx-values. To find this area, we need to set up a definite integral from x=−2x = -2 to x=3x = 3 of the absolute value of the function, since the function is negative and we are interested in the positive area.
  5. Evaluate Lower Limit: The definite integral that represents the enclosed area is ∫−23∣−(x−3)(x+2)∣ dx\int_{-2}^{3} |-(x-3)(x+2)| \, dx.
  6. Calculate Area: Since the function is already negative, taking the absolute value will make it positive. So, the integral simplifies to ∫−23(x−3)(x+2) dx\int_{-2}^{3} (x-3)(x+2) \, dx.
  7. Calculate Area: Since the function is already negative, taking the absolute value will make it positive. So, the integral simplifies to ∫−23(x−3)(x+2) dx\int_{-2}^{3} (x-3)(x+2) \, dx.Now we can calculate the integral. First, we expand the integrand: (x−3)(x+2)=x2−x−6(x-3)(x+2) = x^2 - x - 6.
  8. Calculate Area: Since the function is already negative, taking the absolute value will make it positive. So, the integral simplifies to ∫−23(x−3)(x+2) dx\int_{-2}^{3} (x-3)(x+2) \, dx.Now we can calculate the integral. First, we expand the integrand: (x−3)(x+2)=x2−x−6(x-3)(x+2) = x^2 - x - 6.The integral of x2−x−6x^2 - x - 6 with respect to xx is (1/3)x3−(1/2)x2−6x(1/3)x^3 - (1/2)x^2 - 6x. We will evaluate this antiderivative from −2-2 to 33.
  9. Calculate Area: Since the function is already negative, taking the absolute value will make it positive. So, the integral simplifies to ∫−23(x−3)(x+2) dx\int_{-2}^{3} (x-3)(x+2) \, dx.Now we can calculate the integral. First, we expand the integrand: (x−3)(x+2)=x2−x−6(x-3)(x+2) = x^2 - x - 6.The integral of x2−x−6x^2 - x - 6 with respect to xx is (1/3)x3−(1/2)x2−6x(1/3)x^3 - (1/2)x^2 - 6x. We will evaluate this antiderivative from −2-2 to 33.Plugging in the upper limit of the integral, we get (1/3)(3)3−(1/2)(3)2−6(3)=(1/3)(27)−(1/2)(9)−18=9−4.5−18=−13.5(1/3)(3)^3 - (1/2)(3)^2 - 6(3) = (1/3)(27) - (1/2)(9) - 18 = 9 - 4.5 - 18 = -13.5.
  10. Calculate Area: Since the function is already negative, taking the absolute value will make it positive. So, the integral simplifies to ∫−23(x−3)(x+2) dx\int_{-2}^{3} (x-3)(x+2) \, dx.Now we can calculate the integral. First, we expand the integrand: (x−3)(x+2)=x2−x−6(x-3)(x+2) = x^2 - x - 6.The integral of x2−x−6x^2 - x - 6 with respect to xx is (1/3)x3−(1/2)x2−6x(1/3)x^3 - (1/2)x^2 - 6x. We will evaluate this antiderivative from −2-2 to 33.Plugging in the upper limit of the integral, we get (1/3)(3)3−(1/2)(3)2−6(3)=(1/3)(27)−(1/2)(9)−18=9−4.5−18=−13.5(1/3)(3)^3 - (1/2)(3)^2 - 6(3) = (1/3)(27) - (1/2)(9) - 18 = 9 - 4.5 - 18 = -13.5.Plugging in the lower limit of the integral, we get (1/3)(−2)3−(1/2)(−2)2−6(−2)=(1/3)(−8)−(1/2)(4)+12=−2.666...−2+12=7.333...(1/3)(-2)^3 - (1/2)(-2)^2 - 6(-2) = (1/3)(-8) - (1/2)(4) + 12 = -2.666... - 2 + 12 = 7.333...
  11. Calculate Area: Since the function is already negative, taking the absolute value will make it positive. So, the integral simplifies to ∫−23(x−3)(x+2) dx\int_{-2}^{3} (x-3)(x+2) \, dx.Now we can calculate the integral. First, we expand the integrand: (x−3)(x+2)=x2−x−6(x-3)(x+2) = x^2 - x - 6.The integral of x2−x−6x^2 - x - 6 with respect to xx is (1/3)x3−(1/2)x2−6x(1/3)x^3 - (1/2)x^2 - 6x. We will evaluate this antiderivative from −2-2 to 33.Plugging in the upper limit of the integral, we get (1/3)(3)3−(1/2)(3)2−6(3)=(1/3)(27)−(1/2)(9)−18=9−4.5−18=−13.5(1/3)(3)^3 - (1/2)(3)^2 - 6(3) = (1/3)(27) - (1/2)(9) - 18 = 9 - 4.5 - 18 = -13.5.Plugging in the lower limit of the integral, we get (1/3)(−2)3−(1/2)(−2)2−6(−2)=(1/3)(−8)−(1/2)(4)+12=−2.666...−2+12=7.333...(1/3)(-2)^3 - (1/2)(-2)^2 - 6(-2) = (1/3)(-8) - (1/2)(4) + 12 = -2.666... - 2 + 12 = 7.333...The area is the difference between the antiderivative evaluated at the upper limit and the lower limit: −13.5−(7.333...)=−20.833...-13.5 - (7.333...) = -20.833...

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