Chapter 15.4, Problem 26E

To determine

To find: the area between the curve and the x-axis for the given interval.

Answer to Problem 26E

  $\frac{6}{7}uni{t}^2$ .

Explanation of Solution

Given:

Thegiven intervalis $y=3x^6$ from $x=-1\text{ to }x=1$ .

Calculation:

The area is the integral from $-1\text{ to 1}$ of $3x^6$

Thus, thearea is ${\displaystyle \underset{-1}{\overset{1}{\int }}\left(3x^6\right)dx}$ .

Now, use Fundamental Theorem of Calculus of evaluate. Then,

  $\begin{array}{l}{\displaystyle \underset{-1}{\overset{1}{\int }}\left(3x^6\right)dx}\\ ={\left[3\times \frac{x^{6+1}}{6+1}\right]}_{-1}^1\left\{\because x^{n}dx=\frac{x^{n+1}}{n+1}\right\}\\ ={\left[\frac{3x^7}{7}\right]}_{-1}^1\\ =\left[\frac{3{\left(1\right)}^7}{7}\right]-\left[\frac{3{\left(-1\right)}^7}{7}\right]\\ =\frac{3}{7}-\left[-\frac{3}{7}\right]=\frac{6}{7}\end{array}$

Hence,the area between the curve and the x-axis for the given intervalis $\frac{6}{7}uni{t}^2$ .