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Fild the area of the reglon bounded by the eaphe of f(x)=x^(3)+x^(2)-6x asd f(x)=2x-x^(2)

Fild the area of the reglon bounded by the eaphe of f(x)=x3+x2−6x f(x)=x^{3}+x^{2}-6 x asd f(x)=2x−x2 f(x)=2 x-x^{2}

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Q. Fild the area of the reglon bounded by the eaphe of f(x)=x3+x2−6x f(x)=x^{3}+x^{2}-6 x asd f(x)=2x−x2 f(x)=2 x-x^{2}
  1. Set Equations Equal: Step 11: Set the equations equal to find intersection points.\newlinef(x)=g(x)f(x) = g(x)\newlinex3+x2−6x=2x−x2x^3 + x^2 - 6x = 2x - x^2\newlinex3+2x2−8x=0x^3 + 2x^2 - 8x = 0\newlinex(x2+2x−8)=0x(x^2 + 2x - 8) = 0\newlinex(x+4)(x−2)=0x(x + 4)(x - 2) = 0\newlinex=0,−4,2x = 0, -4, 2
  2. Calculate Definite Integral: Step 22: Calculate the definite integral of the difference between f(x)f(x) and g(x)g(x) from −4-4 to 22.∫−42(x3+x2−6x−(2x−x2)) dx\int_{-4}^{2} (x^3 + x^2 - 6x - (2x - x^2)) \, dx = ∫−42(x3+2x2−8x) dx\int_{-4}^{2} (x^3 + 2x^2 - 8x) \, dx
  3. Integrate the Function: Step 33: Integrate the function.\newline∫(x3+2x2−8x) dx\int(x^3 + 2x^2 - 8x) \, dx\newline= 14x4+23x3−4x2+C\frac{1}{4}x^4 + \frac{2}{3}x^3 - 4x^2 + C
  4. Evaluate the Integral: Step 44: Evaluate the integral from −4-4 to 22.
    [(14)(2)4+(23)(2)3−4(2)2]−[(14)(−4)4+(23)(−4)3−4(−4)2]\left[\left(\frac{1}{4}\right)(2)^4 + \left(\frac{2}{3}\right)(2)^3 - 4(2)^2\right] - \left[\left(\frac{1}{4}\right)(-4)^4 + \left(\frac{2}{3}\right)(-4)^3 - 4(-4)^2\right]
    = [(14)(16)+(23)(8)−4(4)]−[(14)(256)+(23)(−64)−4(16)]\left[\left(\frac{1}{4}\right)(16) + \left(\frac{2}{3}\right)(8) - 4(4)\right] - \left[\left(\frac{1}{4}\right)(256) + \left(\frac{2}{3}\right)(-64) - 4(16)\right]
    = [4+163−16]−[64−1283−64][4 + \frac{16}{3} - 16] - [64 - \frac{128}{3} - 64]
    = −43−(−1283+64)-\frac{4}{3} - (-\frac{128}{3} + 64)
    = −43+1283−64-\frac{4}{3} + \frac{128}{3} - 64
    = 1243−64\frac{124}{3} - 64
    = 1243−1923\frac{124}{3} - \frac{192}{3}
    = −683-\frac{68}{3}

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