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Find the areas of the regions enclosed by the following curves. {:[y=x^(4)-9x^(2)+2" and "y=x^(2)-7],[A=◻" (Type an integer or a simpl) "]:} (Type an integer or a simplified fraction.)

Find the areas of the regions enclosed by the following curves.\newliney=x4−9x2+2 and y=x2−7A=□ (Type an integer or a simpl)  \begin{array}{l} y=x^{4}-9 x^{2}+2 \text { and } y=x^{2}-7 \\ A=\square \text { (Type an integer or a simpl) } \end{array} \newline(Type an integer or a simplified fraction.)

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Q. Find the areas of the regions enclosed by the following curves.\newliney=x4−9x2+2 and y=x2−7A=□ (Type an integer or a simpl)  \begin{array}{l} y=x^{4}-9 x^{2}+2 \text { and } y=x^{2}-7 \\ A=\square \text { (Type an integer or a simpl) } \end{array} \newline(Type an integer or a simplified fraction.)
  1. Identify Intersection Points: Identify the points of intersection of the curves y=x4−9x2+2y = x^4 - 9x^2 + 2 and y=x2−7y = x^2 - 7 by setting them equal to each other.x4−9x2+2=x2−7x^4 - 9x^2 + 2 = x^2 - 7x4−10x2+9=0x^4 - 10x^2 + 9 = 0Let u=x2u = x^2, then u2−10u+9=0u^2 - 10u + 9 = 0(u−9)(u−1)=0(u - 9)(u - 1) = 0u=9u = 9 or u=1u = 1x2=9x^2 = 9 or y=x2−7y = x^2 - 700y=x2−7y = x^2 - 711
  2. Set Up Integral for Area: Set up the integral to find the area between the curves from x=−3x = -3 to x=−1x = -1 and from x=1x = 1 to x=3x = 3.
    Area = ∫−3−1[(x2−7)−(x4−9x2+2)] dx+∫13[(x2−7)−(x4−9x2+2)] dx\int_{-3}^{-1} [(x^2 - 7) - (x^4 - 9x^2 + 2)] \, dx + \int_{1}^{3} [(x^2 - 7) - (x^4 - 9x^2 + 2)] \, dx
    = ∫−3−1[−x4+10x2−9] dx+∫13[−x4+10x2−9] dx\int_{-3}^{-1} [-x^4 + 10x^2 - 9] \, dx + \int_{1}^{3} [-x^4 + 10x^2 - 9] \, dx
  3. Calculate Integral: Calculate the integral.\newlineFor ∫[−x4+10x2−9] dx\int [-x^4 + 10x^2 - 9] \, dx:\newlineAntiderivative = (−1/5)x5+(10/3)x3−9x(-1/5)x^5 + (10/3)x^3 - 9x\newlineEvaluate from −3-3 to −1-1 and from 11 to 33:\newline[(-1/5)(-1)^5 + (10/3)(-1)^3 - 9(-1)] - [(-1/5)(-3)^5 + (10/3)(-3)^3 - 9(-3)]\(\newline+ [(-1/5)(3)^5 + (10/3)(3)^3 - 9(3)] - [(-1/5)(1)^5 + (10/3)(1)^3 - 9(1)]\newline= [(-1/5) + (-10/3) + 9] - [(-243/5) - 30 - 27] + [(243/5) + 30 - 27] - [(-1/5) + (10/3) - 9]\newline= [(-1/5) - 10/3 + 9 + 243/5 + 30 + 27 - 243/5 - 30 + 27 + 1/5 - 10/3 + 9]\newline= [0 + 0 + 36]\newline= \$36\)

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