Chapter 15.4, Problem 36E

To determine

To evaluate: the given definite integrals.

Answer to Problem 36E

  ${\displaystyle \underset{0}{\overset{1}{\int }}\left(x^4+2x^2+1\right)}dx=\frac{28}{15}$ .

Explanation of Solution

Given:

The definite integral is ${\displaystyle \underset{0}{\overset{1}{\int }}\left(x^4+2x^2+1\right)}dx$ .

Calculation:

As thegiven definite integral is ${\displaystyle \underset{0}{\overset{1}{\int }}\left(x^4+2x^2+1\right)}dx$ .

Now, solve the given indefinite integral. Then,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\left(x^4+2x^2+1\right)}dx\\ ={\left[\frac{x^{4+1}}{4+1}-2\times \frac{x^{2+1}}{2+1}+x\right]}_0^1\left\{\because x^{n}dx=\frac{x^{n+1}}{n+1}\right\}\\ ={\left[\frac{x^5}{5}-\frac{2}{3}{x}^3+x\right]}_0^1\\ =\left[\frac{1^5}{5}-\frac{2}{3}{\left(1\right)}^3+1\right]-\left[\frac{0^5}{5}-\frac{2}{3}{\left(0\right)}^3+0\right]\\ =\left[\frac{1}{5}-\frac{2}{3}+1\right]-\left[0\right]=\frac{28}{15}\end{array}$

Hence, the value of the given indefinite integrals is ${\displaystyle \underset{0}{\overset{1}{\int }}\left(x^4+2x^2+1\right)}dx=\frac{28}{15}$ .