ACSEE (Form 6) Integration — practice questions

1. Let $u=\sqrt{x+1}$, so $u^2=x+1$, $x=u^2-1$ and $dx=2u\,du$. 2. The integral becomes $\displaystyle\int\frac{u^2-1}{u}\cdot 2u\,du=\int 2(u^2-1)\,du$. 3. Integrating: $\dfrac{2u^3}{3}-2u+c$. 4. Substituting back $u=\sqrt{x+1}$ gives $\dfrac{2}{3}(x+1)^{3/2}-2\sqrt{x+1}+c$. 5. The sign on the second term is minus because $-1$ was integrated; $\sin^{-1}x$ comes from misreading the denominator as $\sqrt{1-x^2}$. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 9: Integration_

1. Use integration by parts twice, with $\int u\,dv=uv-\int v\,du$. 2. First pass: $u=\theta^2$, $dv=e^{2\theta}d\theta$ gives $\dfrac{\theta^2e^{2\theta}}{2}-\int\theta e^{2\theta}d\theta$. 3. Second pass on $\int\theta e^{2\theta}d\theta$ gives $\dfrac{\theta e^{2\theta}}{2}-\dfrac{e^{2\theta}}{4}$. 4. Combining: $\dfrac{\theta^2e^{2\theta}}{2}-\dfrac{\theta e^{2\theta}}{2}+\dfrac{e^{2\theta}}{4}+c$. 5. Factoring $\dfrac{e^{2\theta}}{4}$ gives $\dfrac{1}{4}e^{2\theta}(2\theta^2-2\theta+1)+c$. The middle sign is minus; a plus there means one parts step dropped its sign. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 9: Integration_

1. For rotation about the $x$-axis, $V=\pi\displaystyle\int_a^b y^2\,dx$. 2. From $y^2-x^3=4$ we get $y^2=4+x^3$. 3. So $V=\pi\displaystyle\int_0^3(4+x^3)\,dx=\pi\left[4x+\frac{x^4}{4}\right]_0^3$. 4. At $x=3$: $12+\dfrac{81}{4}=\dfrac{48+81}{4}=\dfrac{129}{4}$; at $x=0$ the bracket is zero. 5. Hence $V=\dfrac{129}{4}\pi$ cubic units. The value $20.25$ integrates $y^2-x^3$ itself and omits $\pi$ entirely. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 9: Integration_