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If \[y = \begin{cases} \mathrm{k}_1(x) & x \leqslant b \\ \mathrm{k}_2(x) & x \geqslant b \end{cases}\] with \(\mathrm{k}_1(b) = \mathrm{k}_2(b)\), then \(y\) is said to be \emph{continuously differentiable} at \(x = b\) if \(\mathrm{k}_1'(b) = \mathrm{k}_2'(b)\).
The notation \({\lfloor } x \rfloor\) denotes the greatest integer less than or equal to the real number \(x\). Thus, for example, \(\lfloor \pi\rfloor =3\,\), \(\lfloor 18\rfloor =18\,\) and \(\lfloor-4.2\rfloor = -5\,\).
Solution:
The function \(\f\) is defined by $$ \f(x)= \vert x-1 \vert\;, $$ where the domain is \({\bf R}\,\), the set of all real numbers. The function \(\g_n =\f^n\), with domain \({\bf R}\,\), so for example \(\g_3(x) = \f(\f(\f(x)))\,\). In separate diagrams, sketch graphs of \(\g_1\,\), \(\g_2\,\), \(\g_3\,\) and \(\g_4\,\). The function \(\h\) is defined by \[ \h(x) = |\sin {{{\pi}x} \over 2}|, \] where the domain is \({\bf R}\,\). Show that if \(n\) is even, \[ \int_0^n\,\big( \h(x)-\g_n(x)\big)\,\d x = \frac{2n}{\pi} -\frac{n}2\;. \]
Solution:
Show that \[ \int_{-1}^1 \vert \, x\e^x \,\vert \d x =- \int_{-1}^0 x\e^x \d x + \int_0^1 x\e^x \d x \] and hence evaluate the integral. Evaluate the following integrals:
Solution: \begin{align*} && \int_{-1}^1 |x e^x |\d x &= \int_{-1}^0 |xe^x| \d x + \int_0^1 |xe^x| \d x \\ &&&= \int_{-1}^0 -xe^x \d x + \int_0^1 x \e^x \d x \\ &&&= -\int_{-1}^0 xe^x \d x + \int_0^1 x \e^x \d x \\ \\ && \int xe^x \d x &= xe^x - \int e^x \d x \\ &&&= xe^x - e^x \\ \\ \Rightarrow && \int_{-1}^1 |x e^x |\d x &= \left [ xe^x - e^x \right]_0^{-1}+ \left [ xe^x - e^x \right]_0^{1} \\ &&&= -e^{-1}-e^{-1} +e^{0} + e^1 - e^1 +e^0 \\ &&&= 2-2e^{-1} \end{align*}