Question

How do you evaluate the definite integral $\int {{2^x}dx}$ from $\left[ { - 1,1} \right]$ ?

Answer 1

In this question, we are asked to find the definite integral of $\int {{2^x}dx}$ from $\left[ { - 1,1} \right]$.
First we will find the integration of $\int {{2^x}dx}$ using the substitution method and then we will substitute the limits and find the required answer.

Complete step by step answer:
The given integral is $\int\limits_{ - 1}^1 {{2^x}dx}$
First we need to differentiate ${2^x}$ and then solve the definite integral
Let’s start by differentiating ${2^x}$
Let us assume $y = {2^x}$
Taking log on both sides, we get
$\Rightarrow \log y = x\log 2\;$
Differentiating it,
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}x\log 2$
We need to use product rule here,
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}x\log 2$
On rewriting we get,
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{{d(x)}}{{dx}}\log 2 + \dfrac{{d\log 2}}{{dx}}x$
Simplifying it,
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \log 2$
Transferring the variable to the other side we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = y\log 2$
We know that $y = {2^x}$, therefore
$\Rightarrow \dfrac{{dy}}{{dx}} = {2^x}\log 2$
We know that integration is the reverse of differentiating.
If,
$\Rightarrow y = 2x$
$\Rightarrow \dfrac{{dy}}{{dx}} = {2^x}\log 2$
Then,
$\Rightarrow \smallint 2xdx = \dfrac{{{2^x}}}{{\log 2}} + C$
$2x > 0,\forall x \in R$ , there is no negative area so we can insert the limits and evaluate directly.
$\Rightarrow \dfrac{{{2^x}}}{{\log 2}}$
Appling limit from $\left[ { - 1,1} \right]$
$\Rightarrow \dfrac{1}{{\log 2}}({2^1} - {2^{ - 1}})$
On rewriting we get,
$\Rightarrow \dfrac{1}{{\log 2}}\left( {2 - \dfrac{1}{2}} \right)$
Taking LCM on we get,
$\Rightarrow \dfrac{1}{{\log 2}}\left( {\dfrac{3}{2}} \right)$
Then we get,
$\Rightarrow \dfrac{3}{{2\log 2}}$

Therefore the definite integral of $\int {{2^x}dx}$ from $\left[ { - 1,1} \right]$ is $\dfrac{3}{{2\log 2}}$

Note: When such types of questions are asked, if the students aren’t sure about the answer they get, they can check it like this.
We know that integration is the reverse of derivation. For example
The number $2$ is the derivation of $2x$
That means, $2x$is the integration of the number $2$
That is,
$\Rightarrow \dfrac{d}{{dx}}2x = 2$
$\Rightarrow \int {2dx = 2x}$
So when you get the answer, always check it by doing the derivation to the answer you got.
As far this sum, we will check now,
We got the integral of $\int {{2^x}dx}$is $= \dfrac{{{2^x}}}{{\log 2}}$
Taking derivation of the solution,
$\Rightarrow \dfrac{d}{{dx}}\dfrac{{{2^x}}}{{\log 2}} = \dfrac{1}{{\log 2}}\dfrac{d}{{dx}}{2^x}$
We know that,
$\Rightarrow \dfrac{d}{{dx}}{a^x} = {a^x}\log a$
Applying it we get,
$\Rightarrow \dfrac{d}{{dx}}\dfrac{{{2^x}}}{{\log 2}} = \dfrac{1}{{\log 2}}{2^x}\log 2$
Cancelling the log on both numerator and denominator we get,
$\Rightarrow \dfrac{d}{{dx}}\dfrac{{{2^x}}}{{\log 2}} = {2^x}$
Therefore, the answer is correct.