How do you evaluate definite integral $\int {2x - 3} \,dx$ from $\left[ {1,3} \right]$?

Hint: In order to evaluate definite integral $\int {2x - 3} \,dx$ from $\left[ {1,3} \right]$, We will use the formulas of integration like $\int {{x^n} = \dfrac{1}{{n + 1}}{x^{n + 1}}} $ and $\int_a^b {x = \left[ {\dfrac{{{x^2}}}{2}} \right]_a^b = \dfrac{1}{2}\left[ {{b^2} - {a^2}} \right]} $. Thus, by substituting and evaluating, we will determine the required value.

Complete step-by-step answer:
Now, we need to evaluate the definite integral $\int {2x - 3} \,dx$ from $\left[ {1,3} \right]$.
$\int_1^3 {\left( {2x - 3} \right)dx} = \int_1^3 {\left( {2x} \right)dx - \int_1^3 {\left( 3 \right)dx} } $
We know that $\int {{x^n} = \dfrac{1}{{n + 1}}{x^{n + 1}}} $ and $\int_a^b {x = \left[ {\dfrac{{{x^2}}}{2}} \right]_a^b = \dfrac{1}{2}\left[ {{b^2} - {a^2}} \right]} $
Thus, we have,
$ = 2\left[ {\dfrac{{{x^2}}}{2}} \right]_1^3 - 3\left[ x \right]_1^3$
$ = \left( {{3^2} - {1^2}} \right) - 3\left( {3 - 1} \right)$
$ = \left( {9 - 1} \right) - 3\left( 2 \right)$
$ = 8 - 6$
$ = 2$
Hence, $\int_1^3 {\left( {2x - 3} \right)} dx = 2$.
So, the correct answer is “2”.

Note: Integration is a method of adding or summing up the parts to determine the whole. Integration is the calculation of an integral. It is a reverse process of differentiation. This method used to determine the summation under a vast scale and to find useful quantities such as areas, volumes, displacement, etc. The indefinite integrals are used for antiderivatives.