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% !!! IMAGES START HERE !!!

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$\displaystyle \int \frac{dx}{\sqrt{9 + x^2}} $%
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{\newpage\clearpage
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$\displaystyle \int \frac{2x^2-x+1}{(x+1)(x-1)^2} dx $%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline343}%
$\displaystyle
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline345}%
$x = 3 \tan(u) \Rightarrow dx = 3 \sec^2(u) du \ ,
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline347}%
$\displaystyle \int \frac{dx}{\sqrt{9 + x^2}} = \int \frac{3 \sec^2(u) du}
{\sqrt{9 + 9 \tan^2(u)}} = \int \frac{3 \sec^2(u) du}{3 \sec(u)} 
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{\newpage\clearpage
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$C \in {\mathbb{R}}$%
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{\newpage\clearpage
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$\displaystyle \tan(u) = x/3 \Rightarrow \sec(u) = \sqrt{1 + (x/3)^2} =
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline353}%
$\displaystyle \int \frac{dx}{\sqrt{9 + x^2}} = \ln \left|
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{\newpage\clearpage
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$\displaystyle \frac{2x^2 - x + 1}{(x+1)(x-1)^2} = \frac{A}{x+1} +
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline357}%
$\displaystyle 2x^2 - x + 1 = A(x^2 - 2x + 1) + B(x^2-1) + C(x+1)$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline359}%
$\displaystyle 2x^2 - x + 1 = (A + B)x^2 + (C-2A)x + A - B + C$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline361}%
$\displaystyle \left\{ \begin{array}{ccc}
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{\newpage\clearpage
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{\newpage\clearpage
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$\displaystyle 2A = C + 1 = 2 \Rightarrow  A = 1  \ , \
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{\newpage\clearpage
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$\displaystyle \frac{2x^2 - x + 1}{(x+1)(x-1)^2} = \frac{1}{x+1} +
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline369}%
$\displaystyle \int \frac{2x^2 - x + 1}{(x+1)(x-1)^2} dx = 
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline371}%
$\Rightarrow
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline373}%
$\displaystyle \int_5^{\infty} (3x+1)^{-3/2}dx = \lim_{\ell \rightarrow
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline375}%
$\displaystyle \begin{array}{ll}
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline377}%
$\displaystyle \int_5^{\ell} (3x+1)^{-3/2} dx = \int_{16}^{3\ell + 1}
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\right]_{16}^{3 \ell +1}=$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline379}%
$\displaystyle \frac{-2}{3} \left[ \frac{1}{\sqrt{3\ell + 1}} -
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\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline381}%
$\displaystyle \int_5^{\infty} (3x+1)^{-3/2} dx =
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{\newpage\clearpage
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$y^2 = x+4$%
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{\newpage\clearpage
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$y = x+2$%
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{\newpage\clearpage
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$\displaystyle x+4 = (x+2)^2 \Rightarrow x^2 + 4x + 4 = x+ 4 $%
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\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline389}%
$\displaystyle \Rightarrow
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{\newpage\clearpage
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$\displaystyle y_1 = x_1 + 2 = 2 \ , \ y_2 = x_2 + 2 = -1$%
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{\newpage\clearpage
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$x$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline395}%
$\displaystyle A = \int_{-4}^{-3} 2 \sqrt{x+4} dx + \int_{-3}^0 
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\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline397}%
$\displaystyle A = 2 \left[ \frac{(x+4)^{3/2}}{3/2} \right]_{-4}^{-3} +
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\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline399}%
$\displaystyle A = \frac{4}{3} \left( 1^{3/2} - 0^{3/2} \right) +
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\lthtmlinlinemathZ
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline401}%
$\displaystyle A = \frac{4}{3} + \frac{16}{3} - \frac{2}{3} 
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\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline403}%
$y$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline405}%
$\displaystyle y = x + 2 \Rightarrow x = y-2 $%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline407}%
$\displaystyle y^2 = x+4 \Rightarrow x = y^2 - 4$%
\lthtmlinlinemathZ
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline409}%
$\displaystyle A = \int_{-1}^2 \left( y-2 - (y^2 - 4)\right) dy =
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\frac{1}{3} \right) = \frac{9}{2}$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline415}%
$\displaystyle 4x = 5 - x^2 \Rightarrow x^2 + 4x - 5 = (x+5)(x+1) = 0
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline417}%
$\displaystyle V = \pi \int_0^1 \left( (5-x^2)^2 - (4x)^2 \right) dx $%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline419}%
$\displaystyle V = \pi \int_0^4 \left( \frac{y}{4} \right)^2 dy +
\pi \int_4^5 \left( \ \sqrt{5 - y} \ \right)^2 dy =
\frac{\pi}{16} \int_0^4 y^2 dy + \pi \int_4^5 (5-y)dy$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline421}%
$\displaystyle F(x) = \int_{-1}^x f(t) dt \ , \ -1 \leq x \leq 3$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline423}%
$y = f(x)$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline425}%
$F(-1), F(1), F(2), F(3)$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline427}%
$F$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline431}%
$\displaystyle F(-1) = \int_{-1}^{-1} f(t)dt = 0$%
\lthtmlinlinemathZ
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline433}%
$\displaystyle F(1) = \frac{-1}{2} - \frac{1}{2} = -1$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline435}%
$\displaystyle F(2) = F(1) + \frac{1}{2} = \frac{-1}{2} $%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline437}%
$\displaystyle F(3) = F(2) + 1 + \frac{1}{2} = 1$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline439}%
$\displaystyle F'(x) = \frac{d}{dx} \left[ \int_{-1}^x f(t) dt \right] =
f(x) \ , \ x \in (-1,3)$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline441}%
$F'(x)= f(x) < 0 $%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline443}%
$(-1,1) \Longrightarrow F(x) $%
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{\newpage\clearpage
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$(-1,1)$%
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\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline447}%
$F'(x) = f(x) > 0 $%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline449}%
$(1,3) \Longrightarrow F(x) $%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline451}%
$(1,3)$%
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{\newpage\clearpage
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$(4,0)$%
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{\newpage\clearpage
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$(1,0)$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline457}%
$\displaystyle C(4,0) \ , \ F(1,0) \Longrightarrow c = 3 $%
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{\newpage\clearpage
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$X$%
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{\newpage\clearpage
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$C(4,0)$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline463}%
$\displaystyle \frac{(x-4)^2}{a^2} + \frac{(y-0)^2}{b^2} = 1$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline465}%
$(x,y) = (0,0)$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline467}%
$\displaystyle \frac{16}{a^2} + \frac{0^2}{b^2} = 1 \Longrightarrow a = 4$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline469}%
$a^2 = b^2 + c^2$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline471}%
$\displaystyle 4^2 = b^2 + 3^2 \Longrightarrow b^2 = 7$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline473}%
$\displaystyle  \frac{(x-4)^2}{16} + \frac{y^2}{7} = 1$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline475}%
$a,b,c$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}


\end{document}