Question

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

Answer 1

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