How do you integrate \[{{10}^{x}}dx\] from 1 to 0?

Hint: In calculus, an integral is used to find the area under the graph of an equation. Integration, the process of finding an integral, is the reverse of differentiation, the process of finding a derivative. Integration of the exponent function is equal to the exponent divided \[\ln \] of the base of the exponent.
\[\Rightarrow \]\[\int{{{a}^{x}}dx}=\dfrac{{{a}^{x}}}{\ln a}+C\] where C is constant.

Complete step by step answer:
As per the given question we need to integrate \[{{10}^{x}}dx\].
For a closed integral, after integrating the integral we substitute the limits in it.
We first substitute the upper limit and then subtract the lower limit from it.
Now we integrate the given expression. That is
\[\Rightarrow \]\[\int\limits_{1}^{0}{{{10}^{x}}dx}\]
According to the formulae \[\int{{{a}^{x}}dx}=\dfrac{{{a}^{x}}}{\ln a}+C\] if the given formulae has limits then the formula becomes
\[\Rightarrow \]\[\int\limits_{m}^{n}{{{a}^{x}}dx}=\left( \dfrac{{{a}^{x}}}{\ln a} \right)_{m}^{n}=\left( \dfrac{{{a}^{n}}}{\ln a}-\dfrac{{{a}^{m}}}{\ln a} \right)\]
Now according the above formula, the integration of given expression will be
\[\Rightarrow \int\limits_{1}^{0}{{{10}^{x}}dx}=\left( \dfrac{{{10}^{x}}}{\ln 10} \right)_{1}^{0}\]
Now on substituting the limits in the value it becomes
\[\Rightarrow \left( \dfrac{{{10}^{x}}}{\ln 10} \right)_{1}^{0}=\left( \dfrac{{{10}^{0}}}{\ln 10}-\dfrac{{{10}^{1}}}{\ln 10} \right)\]
We know that the value of anything power 0 is 1. That is \[{{a}^{0}}=1\].
\[\Rightarrow \]\[\left( \dfrac{{{10}^{x}}}{\ln 10} \right)_{1}^{0}=\left( \dfrac{1}{\ln 10}-\dfrac{10}{\ln 10} \right)\]
Since the denominator is the same. We can subtract numerators. Then the expression becomes
\[\Rightarrow \left( \dfrac{{{10}^{x}}}{\ln 10} \right)_{1}^{0}=\left( \dfrac{-9}{\ln 10} \right)\]
Therefore, the value of integration of \[{{10}^{x}}dx\] from 1 to 0 is \[\dfrac{-9}{\ln 10}\].

Note:
 In order to solve such a type of problem, we must have enough knowledge on the integrals of different functions. While substituting the limits into the integral check whether we are substituting lower limit or upper limit first. We need to substitute the upper limit first and then subtract the lower limit from it. We should avoid calculation mistakes to get the desired solution.