Use the substitution $u=1+{{x}^{2}}$ to find \(\displaystyle\displaystyle \displaystyle\int{\frac{

Question

Use the substitution $u=1+{{x}^{2}}$ to find $\displaystyle\displaystyle \displaystyle\int{\frac{x{{\mathrm{e}}^{1+{{x}^{2}}}}}{\sqrt{1+{{\mathrm{e}}^{1+{{x}^{2}}}}}}\,}\mathrm{d}x$.[4] Curves ${{C}_{1}}$ and ${{C}_{2}}$ have equations $y=x{{\mathrm{e}}^{{{x}^{2}}-2}}-\frac{1}{2\mathrm{e}}$ and $y=\frac{x{{\mathrm{e}}^{1+{{x}^{2}}}}}{\sqrt{1+{{\mathrm{e}}^{1+{{x}^{2}}}}}}-\frac{1}{2{{\mathrm{e}}^{2}}}$ respectively. The region bounded by the curves ${{C}_{1}}$ and ${{C}_{2}}$, the $y$-axis and the line $x=1$ is $R$. Find the exact area of $R$.[3] The shape of a vase is formed by rotating the part of the curve $x=\sqrt{\frac{2{{y}^{3}}-y+1}{2{{y}^{2}}+1}}$ between $y=0$ and $y=1$ through $2\pi $ radians about the $y$-axis (see diagram below). Find the exact volume of the vase formed.[5]

Accepted Answer

(a)(i) $\sqrt{1+e^{1+x^2}} + C$ (a)(ii) $\sqrt{1+e^2} - \sqrt{1+e}$ units² (b) $\pi\left(\frac{1}{2} + \frac{1}{\sqrt{2}}\tan^{-1}\sqrt{2} - \frac{1}{2}\ln 3\right)$ units³

2022 ASRJC P2 Q3

2022 ASRJC P2 Q3

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