Question

Evaluate the value of integral \[\int{{{a}^{-x}}dx}\].

Answer 1

Hint: To solve this question we will use the formula of integration of \[\int{{{a}^{y}}dy}\] where a is any real number. The number is given as,
\[\int{{{a}^{y}}dy}=\dfrac{{{a}^{y}}}{\log a}+c\]
After proper substitution we will try to obtain \[{{a}^{-x}}\] from \[{{a}^{y}}\] to get our result.

Complete step by step answer:
Given \[\int{{{a}^{-x}}dx}\]
Let \[I=\int{{{a}^{-x}}dx}\]
Let us assume, \[y=-x\].
Differentiating both sides we get,
\[dy=-dx\]
Multiplying by (-1) on both sides we get,
\[-dy=dx\]
Substituting these values in I we get,
\[I=\int{{{a}^{y}}\left( -dy \right)}\]
\[\Rightarrow I=-\int{{{a}^{y}}dy}\] - (1)
Now we finally use a formula of integration stated as,
\[\int{{{a}^{y}}dy}=\dfrac{{{a}^{y}}}{\log a}+c\]
where c is a constant of integration.
\[\Rightarrow \int{{{a}^{y}}dy}=\dfrac{{{a}^{y}}}{\log a}+c\]
Using this in equation (1) we get,
\[I=-\int{{{a}^{y}}dy}\]
\[I=-\dfrac{{{a}^{y}}}{\log a}+c\], where c is constant of integration
 Now replacing \[-x=y\] we get,
\[I=-\dfrac{{{a}^{-x}}}{\log a}+c\]
\[\therefore \] The value of Integral is - \[\dfrac{{{a}^{-x}}}{\log a}+c\], where c is constant of integration

Note:
Students might get confused with \[\int{{{x}^{a}}dx}\] and \[\int{{{a}^{x}}dx}\].
Always remember that,
\[\int{{{x}^{a}}dx}=\dfrac{{{x}^{a+1}}}{a+1}+c\]
And \[\int{{{a}^{x}}dx}=\dfrac{{{a}^{x}}}{\log a}+c\]
Where c is constant of integration