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Integration Questions

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

$$\int_{\mathrm{0}} ^{\:\mathrm{2}} \int_{\mathrm{x}} ^{\:\:\mathrm{3x}\:−\:\mathrm{x}^{\mathrm{2}} } \:\mathrm{6x}^{\mathrm{2}} \:−\:\mathrm{2xy}\:\mathrm{dy}\:\mathrm{dx}\:=\:? \\$$

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

$$\mathrm{32}. \\$$

Commented by A1B1C1D1 last updated on 27/Nov/17

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

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

$${I}\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:\:{A}\left({x}\right){dx}\:\:{with}\:{A}\left({x}\right)\:=\int_{{x}} ^{\mathrm{3}{x}−{x}^{\mathrm{2}} } \left(\mathrm{6}{x}^{\mathrm{2}} −\mathrm{2}{xy}\right){dy}\:\Rightarrow \\$$$${A}\left({x}\right)\:=\mathrm{6}{x}^{\mathrm{2}} \left(\mathrm{3}{x}−{x}^{\mathrm{2}} −{x}\right)\:−\mathrm{2}{x}\:\int_{{x}} ^{\mathrm{3}{x}−{x}^{\mathrm{2}} } {ydy} \\$$$$=\mathrm{12}{x}^{\mathrm{3}} \:−\mathrm{6}{x}^{\mathrm{4}} \:−{x}\left[{y}^{\mathrm{2}} \right]_{{x}} ^{\mathrm{3}{x}−{x}^{\mathrm{2}} } \:=\mathrm{12}{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{4}} \:−{x}\left(\:\:\left(\mathrm{3}{x}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} −{x}^{\mathrm{2}} \right) \\$$$$=\mathrm{12}{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{4}} \:−{x}\left(\:\mathrm{9}{x}^{\mathrm{2}} −\mathrm{6}{x}^{\mathrm{3}} \:+{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \right) \\$$$$=\mathrm{12}{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{4}} \:−{x}^{\mathrm{5}} \:=\mathrm{4}{x}^{\mathrm{3}} \:−{x}^{\mathrm{5}} \:\Rightarrow\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:\left(\mathrm{4}{x}^{\mathrm{3}} −{x}^{\mathrm{5}} \right){dx} \\$$$$=\left[{x}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{6}} \right]_{\mathrm{0}} ^{\mathrm{2}} \:=\mathrm{2}^{\mathrm{4}} −\frac{\mathrm{2}^{\mathrm{6}} }{\mathrm{6}}\:=\mathrm{16}−\frac{\mathrm{16}.\mathrm{4}}{\mathrm{6}}\:=\mathrm{16}\:−\frac{\mathrm{16}.\mathrm{2}}{\mathrm{3}}\:=\frac{\mathrm{48}−\mathrm{32}}{\mathrm{3}}\:=\frac{\mathrm{16}}{\mathrm{3}}\:. \\$$

Answered by ajfour last updated on 27/Nov/17

$$\int_{\mathrm{0}} ^{\:\:\mathrm{2}} \left[\mathrm{6}{x}^{\mathrm{2}} {y}−{xy}^{\mathrm{2}} \right]\mid_{{x}} ^{\mathrm{3}{x}−{x}^{\mathrm{2}} } {dx} \\$$$$=\int_{\mathrm{0}} ^{\:\:\mathrm{2}} \mathrm{6}{x}^{\mathrm{2}} \left(\mathrm{2}{x}−{x}^{\mathrm{2}} \right){dx}−\int_{\mathrm{0}} ^{\:\:\mathrm{2}} {x}\left[\left({x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{2}} \right]{dx}\right. \\$$$$=\left[\mathrm{3}{x}^{\mathrm{4}} −\frac{\mathrm{6}}{\mathrm{5}}{x}^{\mathrm{5}} −\frac{{x}^{\mathrm{6}} }{\mathrm{6}}+\frac{\mathrm{6}}{\mathrm{5}}{x}^{\mathrm{5}} −\mathrm{2}{x}^{\mathrm{4}} \right]\mid_{\mathrm{0}} ^{\mathrm{2}} \\$$$$=\mathrm{16}−\frac{\mathrm{32}}{\mathrm{3}}=\:\frac{\mathrm{16}}{\mathrm{3}}\:. \\$$