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Question Number 24825 by A1B1C1D1 last updated on 27/Nov/17

$${\int }_0^2{\int }_{\mathrm{x}}^{3\mathrm{x}-{\mathrm{x}}^2}{6\mathrm{x}}^2-2\mathrm{xy}\mathrm{dy}\mathrm{dx}=?$$

Commented by 1kanika# last updated on 27/Nov/17

$$32.$$

Commented by A1B1C1D1 last updated on 27/Nov/17

$$\mathrm{Thank}\mathrm{you}.$$

Commented by maxmathsup by imad last updated on 24/May/19

$$I={\int }_0^{2}A\left(x\right)dxwithA\left(x\right)={\int }_x^{3x-x^2}(6x^2-2xy)dy\Rightarrow$$$$A\left(x\right)=6x^2(3x-x^2-x)-2x{\int }_x^{3x-x^2}ydy$$$$=12x^3-6x^4-x{\left[y^2\right]}_x^{3x-x^2}=12x^3-6x^4-x({(3x-x^2)}^2-x^2)$$$$=12x^3-6x^4-x(9x^2-6x^3+x^4-x^2)$$$$=12x^3-6x^4-8x^3+6x^4-x^5=4x^3-x^5\Rightarrow I={\int }_0^2(4x^3-x^5)dx$$$$={[x^4-\frac{1}{6}{x}^6]}_0^2=2^4-\frac{2^6}{6}=16-\frac{16.4}{6}=16-\frac{16.2}{3}=\frac{48-32}{3}=\frac{16}{3}.$$

Answered by ajfour last updated on 27/Nov/17

$${\int }_0^2[6x^{2}y-{xy}^2]{\mid }_x^{3x-x^2}dx$$$$={\int }_0^{2}6x^2(2x-x^2)dx-{\int }_0^{2}x[(x^4-6x^3+8x^2]dx$$$$=[3x^4-\frac{6}{5}{x}^5-\frac{x^6}{6}+\frac{6}{5}{x}^5-2x^4]{\mid }_0^2$$$$=16-\frac{32}{3}=\frac{16}{3}.$$

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