Correct Answer - Option 3 : 0.5
Concept:
Improper integral
Converges: If the limit exists when applied and it is a finite number
Diverges: If a limit does not exist or \(\pm ∞ \)
Calculation:
Given:
Given improper integral \(\displaystyle\int_0^∞ {{\rm{e}}^{ - 2{\rm{t}}}}{\rm{\;dt}}\)
Applying limit by the value a tends to ∞
\(\displaystyle\int_0^\infty {{\rm{e}}^{ - 2{\rm{t}}}}{\rm{\;dt}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{a}} \to \infty } \displaystyle\int_0^{\rm{a}} {{\rm{e}}^{ - 2{\rm{t}}}}{\rm{dt}}\)
\(\displaystyle\int_0^∞ {{\rm{e}}^{ - 2{\rm{t}}}}{\rm{\;dt}}=\mathop {\lim }\limits_{{\rm{a\;}} \to \infty } \left[ {\frac{{{{\rm{e}}^{ - 2{\rm{t}}}}}}{{ - 2}}} \right]\left| {\begin{array}{*{20}{c}} {\rm{a}}\\ 0 \end{array}} \right.\)
\(\displaystyle\int_0^∞ {{\rm{e}}^{ - 2{\rm{t}}}}{\rm{\;dt}}=\left[ { - \frac{1}{2} \times {e^{ - 2\infty }} + \;\frac{1}{2}\times {e^{ - 2\times 0}}} \right]\)
\(\displaystyle\int_0^∞ {{\rm{e}}^{ - 2{\rm{t}}}}{\rm{\;dt}}=0.5\)
∴ The improper integral \(\displaystyle\int_0^∞ {{\rm{e}}^{ - 2{\rm{t}}}}{\rm{dt}}\) converges to 0.5
