0
$\begingroup$

1). Find the function $h(x) = \int_{-1}^{x} f(t)~dt $

$$ f(x)= \begin{cases} \frac{x}{1-2x-x^2}, & x \leq 0 \\ \frac{\sin^3x}{\cos^2x+2\cos x+5}, & x>0 \end{cases} $$

2).Is $h(x)$ Antiderivative / primitive integral of $f(x)$ in domain of $h(x)$?

I found $h(x)$:

$$h(x) = \begin{cases} -\sqrt[]{1-2x-x^2}+\sqrt[]{2}-\arcsin\frac{x+1}{\sqrt2}, & x \in [-1,0] \\ \cos x -2 -\ln|\cos^2x+2\cos x +5| + \ln 8 - 2 \arctan\frac{\cos x+1}{2} +\frac{\pi}{4} +\sqrt[]{2}, & x>0 \end{cases} $$

I have no idea what should I do with the second part of the question. Should I use definition of derivative of $f(x)$ or $h(x)$ in point $x=0$? Or how to prove it without this definition?

  • 1
    2) follows from the fundamental theorem of calculus. – 2017-01-28
  • 0
    So the function F(x) is antidiverative of f(x) when – 2017-01-29
  • 0
    F(x)' =f(x) and F(x) is differentiable in all of it domain? – 2017-01-29
  • 0
    So i have to find if F(x) has a diverative in all of it domain? Do i have to use a definition of diverative in point $x_0 $=0? Or is there other method? – 2017-01-29
  • 1
    What FToC guarantees is as follows: Assume that $I$ is an interval containing $x_0$ and $f : I \to \Bbb{R}$ is continuous. Then $F : I \to \Bbb{R}$ defined by $$ F(x) = \int_{x_0}^{x} f(t) \, dt$$ is an antiderivative of $f$. Notice that it is part of the conclusion that $F$ is differentiable and $F' = f$. Now your function $f$ is continuous on $[-1, \infty)$ and $h(x) = \int_{-1}^{x} f(t) \, dt$, so FToC guarantees that $h$ is differentiable and $h' = f$. – 2017-01-29

1 Answers 1

0

For $x\le0$, $$ \begin{align} \int_{-1}^xf(t)\,\mathrm{d}t &=\int_{-1}^x\frac{t}{1-2t-t^2}\,\mathrm{d}t\\ &=\int_{-1}^x\tfrac1{2\sqrt2}\left(\tfrac{1-\sqrt2}{t+1-\sqrt2}-\tfrac{1+\sqrt2}{t+1+\sqrt2}\right)\,\mathrm{d}t\\ &=\tfrac{1-\sqrt2}{2\sqrt2}\log\left|x+1-\sqrt2\right|-\tfrac{1+\sqrt2}{2\sqrt2}\log\left|x+1+\sqrt2\right|+\tfrac12\log(2)\tag{1} \end{align} $$ For $x\gt0$, $$ \begin{align} \int_{-1}^xf(t)\,\mathrm{d}t &=\tfrac12\log(2)-\tfrac1{\sqrt2}\log\left(\sqrt2+1\right)+\int_0^x\frac{\sin^3(t)}{\cos^2(t)+2\cos(x)+5}\,\mathrm{d}t\\ &=\tfrac12\log(2)-\tfrac1{\sqrt2}\log\left(\sqrt2+1\right)-\int_0^x\frac{1-\cos^2(t)}{\cos^2(t)+2\cos(x)+5}\,\mathrm{d}\cos(t)\\ &=\tfrac12\log(2)-\tfrac1{\sqrt2}\log\left(\sqrt2+1\right)+\int_{\cos(x)}^1\frac{1-t^2}{t^2+2t+5}\,\mathrm{d}t\\ &=\tfrac12\log(2)-\tfrac1{\sqrt2}\log\left(\sqrt2+1\right)+\int_{\cos(x)}^1\left(\frac{2t+6}{t^2+2t+5}-1\right)\,\mathrm{d}t\\ &=\tfrac12\log(2)-\tfrac1{\sqrt2}\log\left(\sqrt2+1\right)+\int_{1+\cos(x)}^2\left(\frac{2t+4}{t^2+4}-1\right)\,\mathrm{d}t\\ &=\tfrac12\log(2)-\tfrac1{\sqrt2}\log\left(\sqrt2+1\right)+\left[\log\left(t^2+4\right)+2\tan^{-1}(t/2)-t\right]_{1+\cos(x)}^2\\ &=\tfrac12\log(2)-\tfrac1{\sqrt2}\log\left(\sqrt2+1\right)+3\log(2)+\frac\pi2-1\\ &-\left(\log\left(\cos^2(x)+2\cos(x)+5\right)+2\tan^{-1}\left(\cos^2(x/2)\right)-\cos(x)\right)\tag{2} \end{align} $$ where $\frac12\log(2)-\frac1{\sqrt2}\log\left(\sqrt2+1\right)$ is $(1)$ evaluated at $x=0$.