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Question Number 192396 by moh777 last updated on 16/May/23

Answered by mehdee42 last updated on 16/May/23

$$4x-x^2=3\Rightarrow x=1,3$$$$v_1=\pi {\int }_0^1{(4x-x^2)}^{2}dx=\pi {\int }_0^1(16x^2-8x^3+x^4)dx=\pi {[\frac{16}{3}{x}^3-2x^4+\frac{x^5}{5}]}_0^1$$$$v_2=18\pi$$$$v_3=\pi {\int }_3^4{(4x-x^2)}^{2}dx=\pi {[\frac{16}{3}{x}^3-2x^4+\frac{x^5}{5}]}_3^4$$$$v=v_1+v_2+v_3$$

Commented by mehdee42 last updated on 17/May/23

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