Chapter 5.5, Problem 2QQ

To determine

To find: The value of an antiderivative $F\left(x\right)$ for $x=8$ .

Answer to Problem 2QQ

The option (B) is correct.

Explanation of Solution

Given data: The function $F\left(x\right)$ is an antiderivative of $f\left(x\right)={\sin }^{3}x$ such that $F\left(1\right)=0$ , then $F\left(8\right)$ .is

  $\left(\text{A}\right)0.373\left(\text{B}\right)0.632\left(\text{C}\right)0.811\left(\text{D}\right)0.968$

Method/Formula used:

The antiderivative $F\left(x\right)$ of the function $f\left(x\right)$ is defined as,

  $F\left(x\right)={\displaystyle \int f\left(x\right)}dx+C$ … (1)

Calculation:

Here, $F\left(x\right)$ is the antiderivative of the function ${\sin }^{3}x$ , therefore, from definition (1),

  $F\left(x\right)={\displaystyle \int {\sin }^{3}x}dx+C$ … (2)

The value of ${\sin }^{3}x$ is given by

  $sin3x=3\sin x-4{\sin }^{3}x$

  ${\sin }^{3}x=\frac{1}{4}\left(3\sin x-sin3x\right)$

Substitute $\frac{1}{4}\left(3\sin x-sin3x\right)$ for ${\sin }^{3}x$ in result (2),

  $\begin{array}{c}F\left(x\right)={\displaystyle \int \frac{1}{4}\left(3\sin x-sin3x\right)}dx+C\\ =\frac{3}{4}{\displaystyle \int \left(\sin x\right)}dx-\frac{1}{4}{\displaystyle \int \left(sin3x\right)}dx+C\leftarrow {\displaystyle \int \left(\sin x\right)}dx=-\cos x\\ =\frac{3}{4}\left[-\cos x\right]-\frac{1}{4}\left[\frac{-\cos 3x}{3}\right]+C\\ =-\frac{3}{4}\cos x+\frac{1}{12}\cos 3x+C\end{array}$

Thus,

  $F\left(x\right)=-\frac{3}{4}\cos x+\frac{1}{12}\cos 3x+C$ … (3)

Substitute 1 for x in result (3),

  $\begin{array}{l}F\left(1\right)=-\frac{3}{4}\cos 1+\frac{1}{12}\cos 3\cdot 1+C\\ F\left(1\right)=-\frac{3}{4}\cos 1+\frac{1}{12}\cos 3+C\leftarrow \cos \left(1\text{rad}\right)=\text{0}\text{.54030231},\cos \left(3\text{rad}\right)=\text{-0}\text{.9899925}\\ F\left(1\right)=-\frac{3}{4}\left(\text{0}\text{.54030231}\right)+\frac{1}{12}\left(\text{-0}\text{.9899925}\right)+C\\ F\left(1\right)=-0.48825+C\end{array}$

But $F\left(1\right)=0$ , therefore, substitute 0 for $F\left(1\right)$ ,

  $\begin{array}{c}F\left(1\right)=-0.48825+C\\ 0=-0.48825+C\\ C=0.48825\end{array}$

Substitute 8 for x and $0.48825$ for C in result (3),

  $\begin{array}{c}F\left(8\right)=-\frac{3}{4}\cos 8+\frac{1}{12}\cos 3\cdot 8+0.48825\\ =-\frac{3}{4}\cos 8+\frac{1}{12}\cos 24+0.48825\leftarrow \cos 8=\text{-0}\text{.14550003,}\cos 24=\text{0}\text{.42417901}\\ =-\frac{3}{4}\left(\text{-0}\text{.14550}\right)+\frac{1}{12}\text{0}\text{.4242}+0.48825\\ =0.633\end{array}$

Therefore, the option (B) is correct.