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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

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