Question

Impropers .... yay, just another page of homework

Answer 1

\[\Large
\begin{array}l
8)\ \int_{0}^{inf}\frac{x}{(x^2+2)^2}dx


\end{array}\]

Answer 2

\[\Large
\begin{array}l
10)\ \int^{-1}_{-inf}\ e^{-2t}dt

\end{array}\]

Answer 3

\[\Large
\begin{array}l
23)\ \int^{inf}_{-inf}\ \frac{x^2}{9+x^6}\ dx

\end{array}\]

Answer 4

\[\Large
\begin{array}l
27)\ \int^{1}_{0}\ \frac{3}{x^5}dx

\end{array}\]

Answer 5

8)\[=\lim_{n \rightarrow \infty}(-1/2)(n^2+2)^{-1}+1/4=1/4\]

Answer 6

these r not difficult....

Answer 7

\[\Large
\begin{array}l
34)\ \int^{1}_{0}\ \frac{1}{4y-1}dy

\end{array}\]

Answer 8

\[\Large
\begin{array}l
40)\ \int^{1}_{0}\ \frac{ln(x)}{\sqrt{x}}dx

\end{array}\]

Answer 9

difficult or not, homework is part of the grade :)

Answer 10

Here's a difficult one :P

Answer 11

I like those curves...

Answer 12

\[\Large
\begin{array}l
8)\ \int_{0}^{inf}\frac{x}{(x^2+2)^2}dx\\\\
=\lim_{t->inf}\ \int_{0}^{t}\frac{x}{(x^2+2)^2}dx\\\\
=\lim_{t->inf}\ \left.-\frac{1}{2(x^2+2)}\right|_{0}^{t}\\\\
=\lim_{t->inf}\ -\frac{1}{2(t^2+2)}+\frac{1}{2(0^2+2)}\\\\
=\frac{1}{4}\\\\

\end{array}\]

Answer 13

waist is a bit too skinny lol