Question

How do you evaluate the indefinite integral $\int{\left( {{x}^{^{2}}}-2x+4 \right)}dx$ ?

Answer 1

Hint: In this question, we have to find the value of an indefinite integral. Thus, we will apply the integration formula and the basic mathematical rules to get the solution. First, we will split the given integral with respect to addition. Then, we will apply the integration formula in all the three integral using the integration formula $\int{{{x}^{m}}dx}=\dfrac{{{x}^{m+1}}}{m+1}$ in the first two integrals and $\int{adx}=ax$ in the last integral. In the end, we will make the necessary calculations, to get the required result for the solution.

Complete step by step solution:
According to the problem, we have to find the value of an indefinite integral.
Thus, we will apply the integration formula and the basic mathematical rules to get the solution.
The indefinite integral given to us is $\int{\left( {{x}^{^{2}}}-2x+4 \right)}dx$ ------------- (1)
First, we will split the integral (1) with respect to addition, we get
$\Rightarrow \int{{{x}^{2}}}dx+\int{-2xdx+\int{4dx}}$
On further simplification the above expression, we get
$\Rightarrow \int{{{x}^{2}}}dx-\int{2xdx+\int{4dx}}$
Now, we will apply the integration formula $\int{{{x}^{m}}dx}=\dfrac{{{x}^{m+1}}}{m+1}$ in first two integrals and $\int{adx}=ax$ in last integral, we get
$\Rightarrow \dfrac{{{x}^{2+1}}}{2+1}-\dfrac{2{{x}^{1+1}}}{1+1}+4x$
On further simplification, we get
$\Rightarrow \dfrac{{{x}^{3}}}{3}-\dfrac{2{{x}^{2}}}{2}+4x$
Now, we know same terms in the numerator and denominator cancel out, thus we get
$\Rightarrow \dfrac{{{x}^{3}}}{3}-{{x}^{2}}+4x+c$ where c is some constant, which is the required answer.
Therefore, the value of integral $\int{\left( {{x}^{^{2}}}-2x+4 \right)}dx$ is $\dfrac{{{x}^{3}}}{3}-{{x}^{2}}+4x$ where c is some constant.

Note:
While solving this problem, do the step-by-step calculation properly to avoid confusion and mathematical error. Mention all the formula you are using to get an accurate answer. Do not forget to write ‘c’ as the constant, because the problem is a type of indefinite integral and not the definite integral.