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Find the area between the curves y=x^(2)+x+1 and y=-2x^(2)-4x-1. Round to 2 decimal places.

1212. Find the area between the curves y=x2+x+1 y=x^{2}+x+1 and y=−2x2−4x−1 y=-2 x^{2}-4 x-1 . Round to 22 decimal places.

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Q. 1212. Find the area between the curves y=x2+x+1 y=x^{2}+x+1 and y=−2x2−4x−1 y=-2 x^{2}-4 x-1 . Round to 22 decimal places.
  1. Set up integral for area: Step 11: Set up the integral for the area between the curves.\newlineTo find the area between two curves, we first need to find the points where they intersect. This is done by setting the equations equal to each other:\newlinex2+x+1=−2x2−4x−1x^2 + x + 1 = -2x^2 - 4x - 1.
  2. Solve for x: Step 22: Solve for x.\newlineCombining like terms, we get:\newline3x2+5x+2=03x^2 + 5x + 2 = 0.\newlineThis can be factored into:\newline(3x+2)(x+1)=0(3x + 2)(x + 1) = 0.\newlineSo, x=−23x = -\frac{2}{3} or x=−1x = -1.
  3. Set up definite integral: Step 33: Set up the definite integral.\newlineThe area AA between the curves from x=−1x = -1 to x=−23x = -\frac{2}{3} is given by the integral of the upper curve minus the lower curve:\newlineA=∫−1−23[(−2x2−4x−1)−(x2+x+1)] dx.A = \int_{-1}^{-\frac{2}{3}} [(-2x^2 - 4x - 1) - (x^2 + x + 1)] \, dx.
  4. Simplify integrand: Step 44: Simplify the integrand.\newlineA=∫−1−23[−3x2−5x−2] dxA = \int_{-1}^{-\frac{2}{3}} [-3x^2 - 5x - 2] \, dx.
  5. Calculate the integral: Step 55: Calculate the integral.\newlineA=[(−x3−52x2−2x)]A = [(-x^3 - \frac{5}{2}x^2 - 2x)] from −1-1 to −23-\frac{2}{3}.\newlineSubstituting the limits:\newlineA=[(−(−23)3−(52)(−23)2−2(−23))−(−(−1)3−(52)(−1)2−2(−1))].A = [(-(-\frac{2}{3})^3 - (\frac{5}{2})(-\frac{2}{3})^2 - 2(-\frac{2}{3})) - (-(-1)^3 - (\frac{5}{2})(-1)^2 - 2(-1))].

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