Solve $\int\limits_1^4 {\left( {3{x^2} + 2x} \right)dx} $?

Hint: In order to find the integration of the given function, we should know what integration is and its types. Integration is nothing but the inverse of differentiation. There are two types of integrals that are as follows: Definite and Indefinite integrals. Indefinite integrals are those that do not have any upper and lower limits, and the one having the lower and upper limits are known as definite integrals.
Formula used:
$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$

Complete answer: We are given an integral function $\int\limits_1^4 {\left( {3{x^2} + 2x} \right)dx} $.
Considering this function to be $I$, that is numerically written as $I = \int\limits_1^4 {\left( {3{x^2} + 2x} \right)dx} $.
Since, we know that there are two separate functions inside the parenthesis, that can be splitted and written as:
$I = \int\limits_1^4 {3{x^2}dx} + \int\limits_1^4 {2xdx} $ ……(1)
Solving each function using the formula from integration, we know that is: $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$.
Using this formula, we get:
$3\int\limits_1^4 {{x^2}dx} = 3\left[ {\dfrac{{{x^{2 + 1}}}}{{2 + 1}}} \right]_1^4 = 3\left[ {\dfrac{{{x^{2 + 1}}}}{3}} \right]_1^4$
Cancelling the common terms, we get:
$ \Rightarrow 3\int\limits_1^4 {{x^2}dx} = \left[ {{x^3}} \right]_1^4$
Since, we have the upper and lower limits, so we substitute the upper limit in the obtained value inside the parenthesis and subtract the value by putting lower limit value from this, and we get:
$ \Rightarrow 3\int\limits_1^4 {{x^2}dx} = \left[ {{4^3} - {1^3}} \right]$
Solving the parenthesis, we get:
$ \Rightarrow 3\int\limits_1^4 {{x^2}dx} = \left[ {64 - 1} \right] = 63$ ……(2)
Similarly, solving the second integral and we get:
$2\int\limits_1^4 {xdx} = 2\left[ {\dfrac{{{x^{1 + 1}}}}{{1 + 1}}} \right]_1^4 = 2\left[ {\dfrac{{{x^2}}}{2}} \right]_1^4$
Cancelling the common terms, we get:
$ \Rightarrow 2\int\limits_1^4 {xdx} = \left[ {{x^2}} \right]_1^4$
Assigning the upper and lower limits and solving them, we get:
$ \Rightarrow 2\int\limits_1^4 {xdx} = \left[ {{4^2} - {1^2}} \right] = \left[ {16 - 1} \right] = 15$ …..(3)

Substituting the values of 2 and 3 in equation 1 and we get:
$
  I = \int\limits_1^4 {3{x^2}dx} + \int\limits_1^4 {2xdx} \\
   \Rightarrow I = 63 + 15 \\
   \Rightarrow I = 78 \\
 $
Therefore, the value of $\int\limits_1^4 {\left( {3{x^2} + 2x} \right)dx} = 78$.

Note:
i.In this integral function, we haven’t given a constant C after integrating, because this is a definite integral and the upper and lower limits are already given, using which the exact value can be found out.
ii.If it was an indefinite integral function, then it was compulsion to have a Constant after integration.