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1.	Solve ∫y=0^{1}∫x=y^{2}^{1}∫z=0^{1-x} x dzdxdy
Answer» \(\displaystyle\int\limits^1_{y=0} \) \(\displaystyle\int\limits^1_{\mathrm x=y^2}\) \(\displaystyle\int\limits^{1-\mathrm x}_{z=0}\) x dz dx dy \(=\displaystyle\int\limits^1_0\left(\int\limits^1_{y^2}\mathrm x \,\left[z\right]^{1-\mathrm x}_0d\mathrm x\right)dy\) \(=\displaystyle\int\limits^1_0\left(\int\limits^1_{y^2}\mathrm x(1-\mathrm x)d\mathrm x\right)dy\) \(=\displaystyle\int\limits^1_0\left(\frac{\mathrm x^2}{2}-\frac{\mathrm x^3}{3}\right)^1_{y^2}dy\) \(=\displaystyle\int\limits^1_0\left(\frac{1}{2}(1-y^4)-\frac{1}{3}(1-y^6)\right)dy\) \(=\frac{1}{2}\displaystyle\int\limits^1_0(1-y^4)dy\) \(-\frac{1}{3}\displaystyle\int\limits^1_0(1-y^6)dy\) \(=\frac{1}{2}\left[y-\frac{y^5}{5}\right]^1_0\) \(-\frac{1}{3}\left[y-\frac{y^7}{7}\right]^1_0\) \(=\frac{1}{2}\left(1-\frac{1}{5}\right)-\frac{1}{3}\left(1-\frac{1}{7}\right)\) \(=\frac{1}{2}\times \frac{4}{5}-\frac{1}{3}\times \frac{6}{7}\) \(=\frac{2}{5}-\frac{2}{7}\) \(=\frac{14-10}{35}=\frac{4}{35}\)

Solve ∫y=0^{1}∫x=y^{2}^{1}∫z=0^{1-x} x dzdxdy

Answer»

\(\displaystyle\int\limits^1_{y=0} \) \(\displaystyle\int\limits^1_{\mathrm x=y^2}\) \(\displaystyle\int\limits^{1-\mathrm x}_{z=0}\) x dz dx dy

\(=\displaystyle\int\limits^1_0\left(\int\limits^1_{y^2}\mathrm x \,\left[z\right]^{1-\mathrm x}_0d\mathrm x\right)dy\)

\(=\displaystyle\int\limits^1_0\left(\int\limits^1_{y^2}\mathrm x(1-\mathrm x)d\mathrm x\right)dy\)

\(=\displaystyle\int\limits^1_0\left(\frac{\mathrm x^2}{2}-\frac{\mathrm x^3}{3}\right)^1_{y^2}dy\)

\(=\displaystyle\int\limits^1_0\left(\frac{1}{2}(1-y^4)-\frac{1}{3}(1-y^6)\right)dy\)

\(=\frac{1}{2}\displaystyle\int\limits^1_0(1-y^4)dy\) \(-\frac{1}{3}\displaystyle\int\limits^1_0(1-y^6)dy\)

\(=\frac{1}{2}\left[y-\frac{y^5}{5}\right]^1_0\) \(-\frac{1}{3}\left[y-\frac{y^7}{7}\right]^1_0\)

\(=\frac{1}{2}\left(1-\frac{1}{5}\right)-\frac{1}{3}\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2}\times \frac{4}{5}-\frac{1}{3}\times \frac{6}{7}\)

\(=\frac{2}{5}-\frac{2}{7}\) \(=\frac{14-10}{35}=\frac{4}{35}\)

Discussion