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Question Number 80951 by TawaTawa last updated on 08/Feb/20

Commented by TawaTawa last updated on 08/Feb/20

$$\mathrm{Please}\mathrm{help}.$$

Commented by john santu last updated on 08/Feb/20

$$\left(1\right)\underset{1}{\stackrel{2}{\int }}\underset{0}{\stackrel{x^2}{\int }}{x}^{2}dydx=$$$$\underset{1}{\stackrel{2}{\int }}{x}^{2}y\underset{0}{\stackrel{x^2}{\mid }}dx=\underset{1}{\stackrel{2}{\int }}\left(x^4\right)dx$$$$=\frac{1}{5}{x}^5\underset{1}{\stackrel{2}{\mid }}=\frac{31}{5}$$

Commented by TawaTawa last updated on 08/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}$$

Answered by mr W last updated on 08/Feb/20

$$\left(1\right)$$$${\int }_1^2{\int }_0^{x^2}{x}^{2}dydx$$$$={\int }_1^2{x}^2{\int }_0^{x^2}dydx$$$$={\int }_1^2{x}^2{\left[y\right]}_0^{x^2}dx$$$$={\int }_1^2{x}^2{x}^{2}dx$$$$={\int }_1^2{x}^{4}dx$$$$=\frac{1}{5}{\left[x^5\right]}_1^2$$$$=\frac{1}{5}(32-1)$$$$=\frac{31}{5}$$$$$$$$\left(2\right)$$$${\int }_0^1{\int }_0^y(x^2+y^2)dxdy$$$$={\int }_0^1{[\frac{x^3}{3}+y^{2}x]}_0^{y}dy$$$$={\int }_0^1(\frac{y^3}{3}+y^{2}y)dy$$$$={\int }_0^1\left(\frac{4y^3}{3}\right)dy$$$$={\left[\frac{y^4}{3}\right]}_0^1$$$$=\frac{1}{3}$$

Commented by TawaTawa last updated on 08/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

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