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Determine whether $\int_{0}^{\pi / 2} \sec x d x$ converges or diverges

$\int_{0}^{\pi / 2} \sec x d x=\ln |\sec x+\tan x| ]_{0}^{\pi / 2}$

But $\sec \left(\frac{\pi}{2}\right)=\infty \quad \Rightarrow$ integral is improper because it is discontinous at $\frac{\pi}{2}$

$\int_{0}^{\pi / 2} \sec x d x=\lim _{t \rightarrow\left(\frac{\pi}{2}\right)^-} \int_{0}^{t} \sec x d x$

$=\lim _{t \rightarrow\left(\frac{\pi}{2}\right)^{-}}|\sec x+\tan x| ]_{0}^{t}$

$\begin{array}{l}{=\lim _{t \rightarrow \frac{\pi}{2}^-}[\ln (\sec t+\tan t)-\ln (1)]=\infty}\end{array}$

The integral is Divergent

Evaluate $\int_{0}^{3} \frac{d x}{x-1}$ if possible

$x-1=0 \quad \rightarrow[x=1]$

$\int_{0}^{3} \frac{d x}{x-1}=\int_{0}^{1} \frac{d x}{x-1}+\int_{1}^{3} \frac{d x}{x-1}$

$\int_{0}^{1} \frac{d x}{x-1}=\lim _{t \rightarrow 1^-} \int_{0}^{t} \frac{d x}{x-1}=\lim _{t \rightarrow 1^-} \ln |x-1| ]_{0}^{t}$

$=\lim _{t \rightarrow {1}^-}[\ln |t-1|-\ln|-1 |]=\lim _{t \rightarrow1^-} \ln (1-t)=-\infty$

The integral is divergent

show that $\int_{0}^{\infty} e^{-x^{2}} d x$ is convergent

$e^{-x} \le e^{-x^{2}} \rightarrow[0,1] \qquad e^{-x^{2}} \leq e^{-x} \rightarrow[1, \infty]$

$\int_{0}^{\infty} e^{-x^{2}} d x-\int_{0}^{1} e^{-x^{2}} d x+\int_{1}^{\infty} e^{-x^{2}} d x$

$\int_{1}^{\infty} e^{-x^{2}} d x$

$x \geq 1 \rightarrow x^{2} \geq x \rightarrow-x^{2} \leq-x$ so that $e^{-x^{2}} \leq e^{-x}$

$\int_{1}^{\infty} e^{-x} d x=\lim _{t \rightarrow \infty} \int_{1}^{t} e^{-x}dx$ and $g(x)=e^{-x^{2}}$

in the comparision theorem we found $\int_{1}^{\infty} e^{-x^{2}} d x$ is convergent so $\int_{0}^{\infty} e^{-x^{2}} d x$

is convergent too