Integration Techniques: Substitution; IBP — ASRJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
- Solve the following integrals.
- \(\displaystyle \int \frac{1-2x}{\sqrt{3+2x-x^2}} \, dx\)
- \(\displaystyle\int x\ln(2-x^2)\,dx\), where \(-\sqrt{2} < x < \sqrt{2}\).
- A function \(f\) is defined by \(f(x) = e^x - 2\).
- Sketch the graph of \(y = |f(x)|\), indicating clearly the equation(s) of asymptote(s), if any.
- Hence find \(\displaystyle\int_0^{\ln 3}|f(x)|\,dx\), giving your answer in exact form.
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(a)(i)
\[\begin{aligned} \int \frac{1-2x}{\sqrt{3+2x-x^2}} \, dx &= \int \frac{2-2x}{\sqrt{3+2x-x^2}} - \frac{1}{\sqrt{2^2-(x-1)^2}} \, dx \\[6pt] &= 2\sqrt{3+2x-x^2} - \sin^{-1}\!\left(\frac{x-1}{2}\right) + c \end{aligned}\] \[ 2\sqrt{3+2x-x^2} - \sin^{-1}\!\left(\frac{x-1}{2}\right) + c \](a)(ii) \(\displaystyle\int x\ln(2-x^2)\,dx\), where \(-\sqrt{2} < x < \sqrt{2}\).
Integration by parts with \(u = \ln(2-x^2)\), \(dv = x\,dx\): \[ \frac{du}{dx} = \frac{-2x}{2-x^2}, \qquad v = \frac{x^2}{2}. \] \[\begin{aligned} \int x\ln(2-x^2)\,dx &= \frac{x^2}{2}\ln(2-x^2) - \int\frac{x^2}{2}\cdot\frac{-2x}{2-x^2}\,dx\\ &= \frac{x^2}{2}\ln(2-x^2) + \int\frac{x^3}{2-x^2}\,dx. \end{aligned}\]
For \(\displaystyle\int\frac{x^3}{2-x^2}\,dx\), perform polynomial division: \(\dfrac{x^3}{2-x^2} = -x + \dfrac{2x}{2-x^2}\).
\[\begin{aligned} \int\frac{x^3}{2-x^2}\,dx &= -\frac{x^2}{2} - \ln|2-x^2| + C. \end{aligned}\] \[ \int x\ln(2-x^2)\,dx = \frac{x^2}{2}\ln(2-x^2) - \frac{x^2}{2} - \ln(2-x^2) + C. \](b)(i) Sketch \(y = |e^x - 2|\).
For \(y = e^x - 2\): y-intercept \((0,-1)\), zero at \(x = \ln 2\), horizontal asymptote \(y = -2\) (for \(x \to -\infty\)). Taking the modulus reflects the portion below the x-axis:
Asymptote: \(y = 2\) (as \(x \to -\infty\)). No asymptote in the positive \(x\) direction.
(b)(ii) \(\displaystyle\int_0^{\ln 3}|e^x-2|\,dx\).
From (i): \(|e^x - 2| = \begin{cases} -e^x + 2 & x \leq \ln 2 \\ e^x - 2 & x > \ln 2 \end{cases}\)
\[\begin{aligned} \int_0^{\ln 3}|e^x-2|\,dx &= \int_0^{\ln 2}(-e^x+2)\,dx + \int_{\ln 2}^{\ln 3}(e^x-2)\,dx\\ &= \bigl[-e^x+2x\bigr]_0^{\ln 2} + \bigl[e^x-2x\bigr]_{\ln 2}^{\ln 3}\\ &= \bigl[(-2+2\ln 2) - (-1+0)\bigr] + \bigl[(3-2\ln 3) - (2-2\ln 2)\bigr]\\ &= (-2+2\ln 2 + 1) + (3 - 2\ln 3 - 2 + 2\ln 2)\\ &= 4\ln 2 - 2\ln 3 - 2 + 1 + 3 - 2\\ &= 4\ln 2 - 2\ln 3. \end{aligned}\] \[ \int_0^{\ln 3}|e^x-2|\,dx = 4\ln 2 - 2\ln 3 = \ln\frac{16}{9} = 2\ln\frac{4}{3}. \]