SOLUTION 4: Compute the area of the region enclosed by the graphs of the equations $ y=x^2 $ and $ y=-x^{2}+4x+6 $ . Begin by finding the points of intersection of the two graphs. From $ y=x^2 $ and $ y=-x^{2}+4x+6 $ we get that $$ x^{2} = -x^{2} + 4x + 6 \ \ \longrightarrow $$ $$ 2x^{2} - 4x - 6 = 0 \ \ \longrightarrow $$ $$ x^{2} - 2x - 3 = 0 \ \ \longrightarrow $$ $$ (x-3)(x+1) = 0 \ \ \longrightarrow \ \ x = 3 \ or \ x = -1 $$ Now see the given graph of the enclosed region.

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Using vertical cross-sections to describe this region, we get that $$ -1 \le x \le 3 \ and \ x^2 \le y \le -x^{2}+4x+6 ,$$ so that the area of this region is $$ AREA = \displaystyle{ \int_{-1}^{3} (Top \ - \ Bottom) \ dx } $$ $$ = \displaystyle { \int_{-1}^{3} ((-x^{2}+4x+6)-(x^{2})) \ dx } $$ $$ = \displaystyle { \int_{-1}^{3} (-2x^{2}+4x+6) \ dx } $$ $$ = \displaystyle { \Big(\frac{-2x^{3}}{3} + \frac{4x^{2}}{2} + 6x \Big) \Big\vert_{-1}^{3} } $$ $$ = \displaystyle { \Big(-\frac{2}{3}x^{3} + 2x^{2} + 6x \Big) \Big\vert_{-1}^{3} } $$ $$ = \displaystyle { \Big( -\frac{2}{3}(3)^{3} + 2(3)^{2} + 6(3) \Big) - \Big( -\frac{2}{3}(-1)^{3} + 2(-1)^{2} + 6(-1) \Big) } $$ $$ = \displaystyle { \Big( -18 + 18 + 18 \Big) - \Big( \frac{2}{3} + 2 - 6 \Big) } $$ $$ = \displaystyle { \Big( 18 \Big) - \Big( \frac{2}{3} - \frac{6}{3} - \frac{18}{3} \Big) } $$ $$ = \displaystyle { \Big( 18 \Big) - \Big( - \frac{22}{3} \Big) } $$ $$ = \displaystyle { \frac{54}{3} + \frac{22}{3} } $$ $$ = \displaystyle {\frac{76}{3}} $$

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