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Exponential integration Quiz 2

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 Evaluate: $\int{{{e}^{x}}{{\left( 3{{e}^{x}}-2 \right)}^{2}}dx.}$

 First substitute the value of $'u'$ in the integration then use the formula of exponential function, $\int{{{e}^{x}}dx=}\text{ }{{e}^{x}}+c$ and $\int{{{x}^{n}}dx=}\dfrac{{{x}^{n+1}}}{n+1}+c$.

  $I=\int{{{e}^{x}}{{\left( 3{{e}^{x}}-2 \right)}^{2}}dx} \\  \text{Let, }3{{e}^{x}}-2=u\Rightarrow du=3{{e}^{x}}dx \\  \Rightarrow I=\dfrac{1}{3}\int{{{u}^{2}}du}=\dfrac{1}{3}\times \dfrac{1}{3}{{u}^{3}}+c \\  \Rightarrow I=\dfrac{1}{9}{{\left( 3{{e}^{x}}-2 \right)}^{3}}+c$

 Evaluate: $\int\limits_{1}^{2}{\dfrac{1}{{{x}^{3}}}}{{e}^{4{{x}^{-2}}}}dx$

 First substitute the value of $'u'$ in the integration then use the formula of exponential function, $\int{{{e}^{x}}dx=}\text{ }{{e}^{x}}+c$.

  $ I=\int\limits_{1}^{2}{\dfrac{1}{{{x}^{3}}}}{{e}^{4{{x}^{-2}}}}dx \\  \text{Let }u={{x}^{-2}}\Rightarrow du=-\dfrac{dx}{3{{x}^{3}}} \\  \text{Limits of the integration, from 1 to }\dfrac{1}{4} \\  \Rightarrow I=-\dfrac{1}{3}\int\limits_{1}^{\dfrac{1}{4}}{{{e}^{4u}}du}=\dfrac{1}{3}\int\limits_{\dfrac{1}{4}}^{1}{{{e}^{4u}}du} \\  \Rightarrow I=\dfrac{1}{12}\left[ {{e}^{4u}} \right]_{\dfrac{1}{4}}^{1}=\dfrac{1}{12}\left( {{e}^{4}}-e \right) $

 Find $q(x)$ if $q'(x)=-0.343{{e}^{-0.49x}}.$

$ q\left( x \right)=\int{-0.343{{e}^{-0.49x\text{ }}}dx} \\  \Rightarrow q\left( x \right)=-0.343\int{{{e}^{-0.49x}}}dx \\  \text{Let, }u=-0.49x\text{ and }du=-0.49dx \\  \Rightarrow q\left( x \right)=\dfrac{-0.343}{-0.49}\int{{{e}^{u}}du}=0.7\int{{{e}^{u}}du} \\  \Rightarrow q\left( x \right)=0.7{{e}^{u}}+c \\  \Rightarrow q\left( x \right)=0.7{{e}^{-0.49x}}+c$

 Evaluate: $\int\limits_{1}^{2}{\dfrac{{{e}^{\dfrac{1}{x}}}}{{{x}^{2}}}dx.}$

 $I=\int\limits_{1}^{2}{\dfrac{{{e}^{\dfrac{1}{x}}}}{{{x}^{2}}}dx} \\  \Rightarrow I=\int\limits_{1}^{2}{{{e}^{{{x}^{-1}}}}{{x}^{-2}}dx} \\  \text{Let }u={{x}^{-1}},du=-{{x}^{-2}}dx \\  \text{Limits of integration becomes, }\dfrac{1}{2}\text{ to 1}\text{.} \\  \text{Now, }I=-\int\limits_{1}^{\dfrac{1}{2}}{{{e}^{u}}du=}\int\limits_{\dfrac{1}{2}}^{1}{{{e}^{u}}du=}\left[ {{e}^{u}} \right]_{\dfrac{1}{2}}^{1} \\  \Rightarrow I=e-{{e}^{\dfrac{1}{2}}}=e-\sqrt{e}$

 Suppose the rate of growth of bacteria in a Petri dish is given by $q\left( t \right)={{3}^{t}}$, where $'t'$ is given in hours and $q(t)$ is given in thousands of bacteria per hour. If a culture starts with $10,000$ bacteria, find a function $Q(t)$ that gives the number of bacteria in the Petri dish at any time $t$. How many bacteria are in the dish after $2$ hours? $\left[ \ln 3=1.098 \right]$

 First find the integration of the given expression, then put the value time in hours for the value of the integral constant then by putting the value for 2 hours.

 $Q(t)=\int{{{3}^{t}}dt=}\dfrac{{{3}^{t}}}{\ln 3}+c \\  \text{Then, at }t=0\text{ we have }Q(0)=10=\dfrac{1}{\ln 3}+c, \\ \text{c}\approx 9.09\text{ so, we get;} \\  Q(t)=\dfrac{{{3}^{t}}}{\ln 3}+9.09 \\  \text{At time }t=2\text{, }Q(2)=\dfrac{{{3}^{2}}}{\ln 3}+9.09=17.282 \\  \text{After 2 hours, there are 17,282 bacteria in the dish}$