2 problems found
If \({\rm f}(t)\ge {\rm g}(t)\) for \(a\le t\le b\), explain very briefly why \(\displaystyle \int_a^b {\rm f}(t) \d t \ge \int_a^b {\rm g}(t) \d t\). Prove that if \(p>q>0\) and \(x\ge1\) then $$\frac{x^p-1}{ p}\ge\frac{x^q-1}{ q}.$$ Show that this inequality also holds when \(p>q>0\) and \(0\le x\le1\). Prove that, if \(p>q>0\) and \(x\ge0\), then $$\frac{1}{ p}\left(\frac{x^p}{ p+1}-1\right)\ge \frac{1}{q}\left(\frac{x^q}{ q+1}-1\right).$$
Solution: This is just the result that all of the area beneath \(g(t)\) is also below \(f(t)\) If \(p > q > 0, x \geq 1 \Rightarrow x^p \geq x^q\), therefore applying the result we have \begin{align*} && \int_1^x x^p\, \d t & \geq \int_1^x x^q\, \d t \\ \Rightarrow && \frac{x^p-1}{p} & \geq \frac{x^q-1}{q} \end{align*} When \(p > q > 0, 0 \leq x \leq 1\) we have \(x^p \leq x^q\), ie \begin{align*} && \int_x^1 x^q\, \d t & \geq \int_{x}^1 x^p\, \d t \\ \Rightarrow && \frac{1-x^q}{q} & \geq \frac{1-x^p}{p} \\ \Rightarrow && \frac{x^p-1}{p} &\geq \frac{x^q-1}{q} \end{align*} Now looking at the functions \(f(x) = \frac{x^p-1}{p}, g(x) = \frac{x^q-1}{q}\) and \(x \geq 1\) we have \begin{align*} && \int_0^x \frac{t^p-1}{p} \d t & \geq \int_0^x \frac{t^q-1}{q} \d t\\ \Rightarrow &&\frac1p \left[\frac{t^{p+1}}{p+1} - t \right]_0^x &\geq\frac1q \left[\frac{t^{q+1}}{q+1} - t \right]_0^x \\ \Rightarrow &&\frac1p \left(\frac{x^{p+1}}{p+1} -x\right) &\geq\frac1q \left(\frac{x^{q+1}}{q+1} - x \right)\\ \Rightarrow &&\frac1p \left(\frac{x^{p}}{p+1} -1\right) &\geq\frac1q \left(\frac{x^{q}}{q+1} - 1 \right)\\ \end{align*}
Given that \(a\) is constant, differentiate the following expressions with respect to \(x\):
Solution: \begin{align*} && y &= x^a \\ && \frac{\d y}{\d x} &= \begin{cases} ax^{a-1} & a \neq 0 \\ 0 & a = 0 \end{cases} \\ \\ && y &= a^x \\ &&&= e^{(\ln a) \cdot x} \\ && \frac{\d y}{\d x} &= \ln a e^{(\ln a) x} \\ &&&= \ln a \cdot a^ x \\ \\ && y &= x^x \\ &&&= e^{x \ln x}\\ && \frac{\d y}{\d x} &= e^{x \ln x} \cdot \left ( \ln x + x \cdot \frac1x \right) \\ &&&= x^x \left (1 + \ln x \right) \\ \\ && y&= x^{(x^x)} \\ &&&= e^{x^ x \cdot \ln x} \\ && \frac{\d y}{\d x} &= e^{x^x \cdot \ln x} \left ( x^x \left (1 + \ln x \right) \cdot \ln x + x^x \cdot \frac1x\right) \\ &&&= x^{x^x} \left (x^x (1+ \ln x) \ln x +x^{x-1} \right) \\ &&&= x^{x^x+x-1} \left (1 + x \ln x + x (\ln x)^2 \right) \\ \\ && y &= (x^x)^x \\ &&&= x^{2x} \\ &&&= e^{2x \ln x} \\ && \frac{\d y}{\d x} &= e^{2 x \ln x} \left (2 \ln x + 2 \right) \\ &&&= 2(x^x)^x(1 + \ln x) \end{align*}