Chapter 13, Problem 1RE

To determine

To calculate: The solution of indefinite integral ${\displaystyle \int \left(x^2-8x-7\right)dx}$.

Answer to Problem 1RE

Solution:

The solution of indefinite integral is $\underset{\_}{{\displaystyle \int \left(x^2-8x-7\right)dx}=\frac{x^3}{3}-4x^2-7x+c}$.

Explanation of Solution

Given:

The indefinite integral is ${\displaystyle \int \left(x^2-8x-7\right)dx}$.

Formula used:

$\begin{array}{ccc}\hfill {\displaystyle \int f\left(x\right)}dx& =& {\displaystyle \int x^n}dx\hfill \\ \hfill & =& \frac{x^{n+1}}{n+1}+c\hfill \end{array}$

Sum or Difference Rule is;

${\displaystyle \int \left[f\left(x\right)\pm g\left(x\right)\right]}dx={\displaystyle \int f\left(x\right)}dx\pm {\displaystyle \int g\left(x\right)dx}$

Calculation:

The integration of function $f\left(x\right)=x^n$ for $n\ne -1$ is;

$\begin{array}{ccc}\hfill {\displaystyle \int f\left(x\right)}dx& =& {\displaystyle \int x^n}dx\hfill \\ \hfill & =& \frac{x^{n+1}}{n+1}+c\hfill \end{array}$

Here $c$ is an arbitrary constant.

So, integration can be calculated as;

$\begin{array}{ccc}\hfill {\displaystyle \int \left(x^2-8x-7\right)dx}& =& \frac{x^3}{3}-\frac{8x^2}{2}-7x+c\hfill \\ \hfill & =& \frac{x^3}{3}-4x^2-7x+c\hfill \end{array}$