Evaluate $\int_{1}^{3}{\left( {{x}^{2}}+3x+{{e}^{x}} \right)}dx$.

Hint: The integral of sum of two or more than two functions is equal to the sum of integrals of each function, that is, $\int{\left[ f\left( x \right)+g\left( x \right)dx \right]}=\int{f\left( x \right)}dx+\int{g\left( x \right)}dx$. The integration is mainly of two types, first is definite and second is indefinite. The integral which contains upper and lower limit is called a definite integral and which does not contain upper and lower limits are called as indefinite integral. Here, in this question the definite integral is given as the upper and lower limit is present in this function.

Complete step by step solution:
 Let $I=\int_{1}^{3}{\left( {{x}^{2}}+3x+{{e}^{x}} \right)}dx$
$\Rightarrow I=\int_{1}^{3}{{{x}^{2}}dx}+\int_{1}^{3}{3x}dx+\int_{1}^{3}{{{e}^{x}}dx}\ldots .\left( 1 \right)$
Now, we calculate $\int_{1}^{3}{{{x}^{2}}dx}$ and by $\int_{1}^{3}{3xdx}$ using the formula $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}:-$
$\int_{1}^{3}{{{x}^{2}}dx}=\left[ \dfrac{{{x}^{3}}}{3} \right]_{1}^{3}=\dfrac{1}{3}\left[ {{\left( 3 \right)}^{3}}-{{\left( 1 \right)}^{3}} \right]=\dfrac{1}{3}\left( 27-1 \right)=\dfrac{26}{3}$
$\int_{1}^{3}{3xdx}=3\int_{1}^{3}{xdx}=3\left[ \dfrac{{{x}^{2}}}{2} \right]_{1}^{3}=\dfrac{3}{2}\left[ {{\left( 3 \right)}^{2}}-{{\left( 1 \right)}^{2}} \right]=\dfrac{3}{2}\left( 9-1 \right)=\dfrac{3}{2}\left( 8 \right)=12$
Now we use the formula $\int{{{e}^{x}}dx={{e}^{x}}}$ to calculate $\int_{1}^{3}{{{e}^{x}}dx}:-$
$\int_{1}^{3}{{{e}^{x}}dx}=\left[ {{e}^{x}} \right]_{1}^{3}={{e}^{3}}-{{e}^{1}}={{e}^{3}}-e$
Now, putting all the obtained values in (1), we get
$I=\dfrac{26}{3}+12+{{e}^{3}}-e$
$\Rightarrow I=\dfrac{26+36}{3}+{{e}^{3}}-e$
$\Rightarrow I=\dfrac{62}{3}+{{e}^{3}}-e$

Hence,$\int_{1}^{3}{\left( {{x}^{2}}+3x+{{e}^{x}} \right)}dx=\dfrac{62}{3}+{{e}^{3}}-e$

Note: In this question we have to solve the integration for given limits. As we know that integration is nothing but summation of discrete data or the integrals. We can also say that it is the reverse process of differentiation.
Let’s take an example for better understanding; we know the derivative of is \[\cos x\]. We can write in mathematical form as, \[\left( \dfrac{d}{dx} \right)\sin x=\cos x\] so, the differentiation of \[\sin x\] is \[\cos x\], thus \[\sin x\] is the antiderivative of \[\cos x\] that is, it is the integration of the \[\cos x\].
The general form of representing an integral function is,
\[y=\int{\text{f}(x)\text{ }dx}\]
\[\int{\text{f}(x)\text{ }dx}=\text{F}(x)+\text{C}\]
Where,\[f\] is the anti-derivative or integration of the function, \[\int{{}}\] is the symbol of integral and \[\text{C}\] is the constant term. Always remember that the integration of constant terms is always zero.
Use the integration formulas $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$ and $\int{{{e}^{x}}dx={{e}^{x}}}$ to evaluate the values of $\int_{1}^{3}{{{x}^{2}}dx}$ , $\int_{1}^{3}{3xdx}$ and $\int_{1}^{3}{{{e}^{x}}dx}$ respectively.