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 \centerline{\bf {\underbar {Solutions} to Mathematics 166, Exam 1}}
\centerline{\bf Spring Semester, 1992 }
\centerline{\bf {February  17, 1992}} 
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\noindent
{\bf1.}  Evaluate each of the following integrals.  \newline
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\noindent
a) $\int xe^{x^{2}} dx = \frac {1}{2} e^{x^2} + C$ 

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\noindent
b)  $\int {\frac {\sin 2x}{10+ \cos 2x}} dx = -\frac{1}{2} \int \frac {d(10 + \cos
2x)}{10 + \cos 2x} = -\frac {1}{2} ln (10 + \cos 2x) + C$ 

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\noindent
c)  $\int {\frac{x}{x - 1}} dx = \int (1 + \frac{1}{x-1}) dx = x + ln |x-1| + C$

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\noindent
d)  $\int \sin (\cos 2x) \sin 2x dx = -\frac {1}{2} \int \sin (\cos 2x) d (\cos 2x) =
\frac {1}{2} \cos (cos 2x) + C $

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\noindent
e)  $\int_0^{\frac 14} {\frac {1}{\sqrt{1 - 4x^2}}dx} = \frac {1}{2} \int_0^{\frac 14}
\frac{ d(2x) }{\sqrt{1-(2x)^2}}= \frac 12 \sin^{-1}(2x) |_0^{\frac 14} = \frac 12
(\frac{\pi}{6} - 0) = \frac {\pi}{12}$ 
 
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\noindent
f)  $\int {\frac{3x^2 + 6}{x^3 + 6x + 5}} dx = \int \frac {d(x^3 + 6x + 5)}{x^3 + 6x + 5} =
ln |x^3 + 6x + 5| + C$
   

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\noindent
g)  $\int {(\sqrt{x} + \frac {2}{\sqrt{x}})} dx = \int x^{\frac 12} dx + 2 \int
x^{ -\frac{1}{2}} dx = \frac {x^{\frac {1}{2}} + 1}{\frac {1}{2} + 1} + 2 \frac {x^{ -\frac
{1}{2}} + 1}{-\frac {1}{2} + 1} + C = \frac 23 x^{\frac {3}{2}} + 4x^{\frac 12} + C$
  

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\noindent
h)  $\int_{0}^{\pi^2} \frac{\sin{\sqrt {x}}}{{\sqrt {x}}} dx = 2 \int \sin \sqrt {x}  \;\; d(
\sqrt{x}) = -2(\cos \sqrt {x}) |_{0}^{\pi^2} = -2 (-1-1) = 4$ 

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\noindent
{\bf 2.}  
 We have that \newline

$I= \int {\frac{2x + 2}{(x^2 + 2x + 5)^2}} dx +
\int{\frac{1}{(x^2 + 2x + 5)^2}} dx$ 
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\noindent
$=\frac{(x^2 + 2x + 5)^{-2+1}}{-2 + 1} + \int \frac{1}{[(x+1)^2 + 4]^2} dx$ \newline
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\noindent
$=\frac{-1}{x^2 + 2x + 5} + \int {\frac{1}{4^2[(\frac{x + 1}{2})^2 + 1]^2}}dx$
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\noindent
(Let $\frac{x + 1}{2} = t.)$
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\noindent
$= \frac{-1}{x^2 + 2x + 5} + \frac{2}{4^2} \int{\frac{1}{(t^2 + 1)^2}}dt$ \newline
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\noindent
$=\frac{-1}{x^2 + 2x + 5} + \frac{1}{8} [\frac12 \frac{t}{t^2 +1} + \frac 12 \int
\frac {1}{t^2 + 1}dt]$
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\noindent
$= \frac{-1}{x^2 + 2x + 5} + \frac {1}{16} \frac{\frac{x + 1}{2} }{(\frac{x+1}{2})^2 + 1}
+ \frac{1}{16} \tan^{-1} (\frac{x+1}{2}) + C.$
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\noindent
{\bf 3.}  We have:
$$\frac {1}{x^2(x^2 + 1)} = \frac {A}{x} + \frac {B}{x^2} + \frac {Cx + D}{x^2 + 1}$$
\noindent
Therefore  \newline
$1 = Ax(x^2 + 1) + B(x^2 + 1) + (Cx + D)x^2$ \newline
or $1 = (A + C)x^3 + (B + D)x^2 + Ax + B$ \newline
or $B = 1, \;\; A=0, \;\; C = 0, \;\; D = -1 $
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\noindent
Thus  \newline
$I = \displaystyle \lim_{b\to\infty} \left[ \int_1^b \frac{dx}{x^2} - \int_1^b
\frac{dx}{x^2 + 1} \right] 
= \displaystyle \lim_{b\to\infty} \left[ -x^{-1} - tan^{-1}x \right]_{1}^{b}$
\newline
$ = \displaystyle \lim_{b\to\infty} \left[ -b^{-1} + 1 - tan ^{-1}b + tan ^{-1}1 \right]
 = 1 - \frac{\pi}{2} + \frac{\pi}{4} = 1 - \frac {\pi}{4}.$

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\noindent
{\bf4.} 
If we multiply by $e^{-x^4}$ we obtain $ \left( ye^{-x^4} \right)' =
x^3e^{-x^4}$  \newline
\noindent
Thus,  $ye^{-x^4} = \int {x^3e^{-x^4}} dx + C = y(x) = \frac14 + ce^{x^4}.$
\newline
\noindent
Since $-1 = y(0) = - \frac 14  + C,$ we have $C = -\frac {3}{4}$
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\noindent
and $\boxed {y(x) = \frac {-1}{4} (1 + 3e^{x^4})}.$
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\noindent
{\bf  5.}  We have $ \displaystyle {\int x^ne^{-x}} dx = \displaystyle {\int x^nd(-e^{-x})}
dx = -x^ne^{-x} + \displaystyle {\int e^{-x}nx^{n-1}}$
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\noindent
$=-x^ne^{-x} + n {\int x^{n-1}e^{-x}} dx.$
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\noindent
Next  $ \displaystyle{\int_{0}^{\infty} x^n e^{-x}  } dx = -x^n e^{-x} |_{0}^{\infty} + n
{\int_{0}^{\infty} x^{n-1}e^{-x}}dx $

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\noindent
Since $\displaystyle \lim_{x\to\infty} x^ne^{-x}  = 0$ we have \newline
$\displaystyle {\int_0^\infty x^ne^{-x}} dx = n {\int_0^\infty x^{n-1}e^{-x}dx}.$
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\noindent
By repeating the above argument, we obtain  \newline
${\int_{0}^{\infty} x^ne^{-x}}dx =
n(n-1)(n-2) \cdots 1 \cdot {\int_0^{\infty}} e^{-x} dx = n!$

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\noindent
since $\int_0^{\infty} e^{-x} dx = 1.$ 


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