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Question Number 201702 by hardmath last updated on 10/Dec/23

$$\mathrm{Find}:$$$${\int }_1^3\mathrm{dx}{\int }_{\mathbf{x}}^{{\mathbf{x}}^3}(\mathrm{x}-\mathrm{y})\mathrm{dy}$$

Commented by mr W last updated on 11/Dec/23

$$takecare!ithinkyouwanttomean$$$${\int }_1^3{\int }_x^{x^3}(x-y)dydx$$

Answered by mr W last updated on 11/Dec/23

$${\int }_1^3{\int }_x^{x^3}(x-y)dydx$$$$={\int }_1^3\left[{\int }_{x^3}^x(x-y)d(x-y)\right]dx$$$$={\int }_1^3{\left[\frac{{(x-y)}^2}{2}\right]}_{x^3}^{x}dx$$$$=-{\int }_1^3\frac{{(x-x^3)}^2}{2}dx$$$$=-\frac{1}{2}{\int }_1^3(x^2-2x^4+x^6)dx$$$$=-\frac{1}{2}{[\frac{x^3}{3}-\frac{2x^5}{5}+\frac{x^7}{7}]}_1^3$$$$=-\frac{1}{2}(\frac{3^3-1^3}{3}-\frac{2(3^5-1^5)}{5}+\frac{3^7-1^7}{7})$$$$=-\frac{1}{2}(\frac{26}{3}-\frac{2\times 242}{5}+\frac{2186}{7})$$$$=-\frac{11768}{105}$$

Commented by hardmath last updated on 11/Dec/23

$$\mathrm{cool}\mathrm{dear}\mathrm{professor}$$

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