find the area bounded by the curve y=3x-x^3 and the line y=2
Area enclosed:
(3×1)−∫−2−3(3x−x3)dx−∫01(3x−x3)dx+∣∫−30(3x−x3)dx∣⇒3−[3x22−x44]−2−3−[3x22−x44]01+∣[3x22−x44]−30∣⇒3−[92−94−6+4]−[32−14−0+0]+∣(0−0)−(92−94)∣⇒3−14−54+94⇒154 sq.units(3\times1)-\int_{-2}^{-\sqrt{3}}(3x-x^3)dx-\int^1_0(3x-x^3)dx+|\int^0_{-\sqrt{3}}(3x-x^3)dx|\\ \Rightarrow3-[\frac{3x^2}{2}-\frac{x^4}{4}]^{-\sqrt{3}}_{-2}-[\frac{3x^2}{2}-\frac{x^4}{4}]^{1}_{0}+|[\frac{3x^2}{2}-\frac{x^4}{4}]_{-\sqrt{3}}^{0}|\\ \Rightarrow3-[\frac{9}{2}-\frac{9}{4}-6+4]-[\frac{3}{2}-\frac{1}{4}-0+0]+|(0-0)-(\frac{9}{2}-\frac{9}{4})|\\ \Rightarrow 3-\frac{1}{4}-\frac{5}{4}+\frac{9}{4}\\ \Rightarrow \frac{15}{4} \ sq. units(3×1)−∫−2−3(3x−x3)dx−∫01(3x−x3)dx+∣∫−30(3x−x3)dx∣⇒3−[23x2−4x4]−2−3−[23x2−4x4]01+∣[23x2−4x4]−30∣⇒3−[29−49−6+4]−[23−41−0+0]+∣(0−0)−(29−49)∣⇒3−41−45+49⇒415 sq.units
