Question

How do you evaluate the definite integral $\int {\dfrac{{dx}}{x}} $ from $[1,e]?$

Answer 1

Hint: In this question, we want to find the value of the definite integral. The final value of a definite integral is the value of integral to the upper limit minus the value of the definite integral for the lower limit.
$\int\limits_a^b {f(x)dx = F(b) - F(a)} $
In the above equation,
F(b) and F(a) are the integral functions for the upper and lower limit respectively, f(x) is the integrand, dx is the integrating agent, ‘b’ is the upper limit of the definite integral, and ‘a’ is the lower limit of the definite integral.

Complete step-by-step solution:
In this question, we want to find the definite integral $\int {\dfrac{{dx}}{x}} $ from [1,e].
Here, the interval is given as [1,e]. That means the upper limit is e and the lower limit 1.
So, we can write the given integral as:
 $ \Rightarrow \int\limits_1^e {\dfrac{1}{x}} dx$
Now, by the fundamental theorem of calculus, we get: $\dfrac{d}{{dx}}(\ln x) = \dfrac{1}{x}$ .
So, the anti-derivative of $\dfrac{1}{x}$ is $\ln \left| x \right|$ .
Therefore,
$ \Rightarrow \int\limits_1^e {\dfrac{1}{x}} dx = \left[ {\ln x} \right]_1^e$
Let us apply the limit in the above expression.
So,
$ \Rightarrow \int\limits_1^e {\dfrac{1}{x}} dx = \ln e - \ln 1$
Now, as we know that the value of $\ln e$ is equal to 1 and the value of $\ln 1$ is equal to 0.
Let us substitute these values in the above expression.
$ \Rightarrow \int\limits_1^e {\dfrac{1}{x}} dx = 1 - 0$
Let us simplify the right-hand side.
$ \Rightarrow \int\limits_1^e {\dfrac{1}{x}} dx = 1$

Hence, the value of the definite integral $\int {\dfrac{{dx}}{x}} $ from $[1,e]$ is $1.$

Note: Indefinite integrals: Indefinite integrals are those integrals that do not have any limit of integration. Indefinite integral has an arbitrary constant.
Definite integrals: Definite integrals are those integrals that have an upper and lower limit. Define integrals have two different values for the upper limit and lower limit when they are evaluated. The final value of a definite integral is the value of integral to the upper limit minus the value of the definite integral for the lower limit.