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Find the area of the region bounded by y = x4 - 3x3 - 4x2 + 10, y = 40 - x2, x = 1 and x = 3. |
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Answer» The required area of region is bounded above by y = 40 - x2 and bounded below by y - x4 - 3x3 - 4x2+ 10 and between the lines x = 1 and x = 3 \(\therefore\) Required area = \(\int\limits_1^3[(40-x^2)+-(x^4-3x^3-4x^2+10)]dx\) = \(\int\limits_1^3(-x^4+3x^3+3x^2+30)dx\) = \([-\frac{x^5}5+\frac{3x^4}4+x^3+30x]_1^3\) = \(\frac{-1}5\)(35 - 1) + \(\frac{3}4\)(34 - 1) + (33 - 1) + 30(3 - 1) = \(\frac{-1}5\) (243 - 1) + \(\frac{3}4\)(81 - 1) + (27 - 1) + 30 x 2 = \(\frac{-1}5\) x 242 + 3 x 20 + 26 + 60 = 146 - \(\frac{242}5\) = \(\frac{488}5\) square units. |
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