A = \left[ 2x^2 - \frac{2}{3}x^3 \right]_0^2 = \left( 2(2)^2 - \frac{2}{3}(2)^3 \right) - \left( 2(0)^2 - \frac{2}{3}(0)^3 \right).
![A = \left[ 2x^2 - \frac{2}{3}x^3 \right]_0^2 = \left( 2(2)^2 - \frac{2}{3}(2)^3 \right) - \left( 2(0)^2 - \frac{2}{3}(0)^3 \right).](https://soloferat.biz.id/images/a--left-2x2---frac23x3-right02--left-222---frac2323-right---left-202---frac2303-right.jpg)
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- The area \( A \) between the curves from \( x = 0 \) to \( x = 2 \) is given by:
- A = \int_0^2 \left((4x - x^2) - x^2\right) \, dx = \int_0^2 (4x - 2x^2) \, dx.
- Evaluating the integral:
- Calculating each term:
- = \left( 8 - \frac{16}{3} \right) = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}.
- \boxed{\frac{8}{3}}.
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