Chapter 8.2, Problem 49E

To determine

To Find: The positive value of k

Answer to Problem 49E

The positive value of k$\approx 1.8269$

Explanation of Solution

Given information:

The area of the region enclosed between the graph of $y=k\cos x$ and the graph of $y=k{x}^2$ is $2$

Concept used: If $f(x)$ and $g(x)$ are continuous with $f(x)\ge g(x)$ throughout $[a,b]$ , then the area between the curves $y=f(x)$ and $y=g(x)$ from a to b is the integral of $[f(x)-g(x)]$ from a to b

  $A={\displaystyle \underset{a}{\overset{b}{\int }}[f(x)-g(x)]}dx$

Calculation:

The area of the region enclosed between the graph of $y=k\cos x$ and the graph of $y=k{x}^2$ is $2$

Two graphs

  $y=k\cos x,y=k{x}^2$

Using a calculator, solve the equation

  $\cos x=x^2$

  

The solution are $-0.8241$ and $0.8241$

graph of $y=k\cos x$ is above the graph of $y=k{x}^2$

Therefore, the area is

  $A={\displaystyle \underset{-0.8241}{\overset{0.8241}{\int }}\left[k\cos x-k{x}^2\right]}dx$

  $"k"$ is a constant

  $\begin{array}{l}A=k{\displaystyle \underset{-0.8241}{\overset{0.8241}{\int }}\left[\cos x-x^2\right]}dx\\ A=k{\left[\sin x-\frac{x^3}{3}\right]}_{-0.8241}^{0.8241}\\ A=k\left[\sin 0.8241-\frac{{\left(0.8241\right)}^3}{3}+\sin 0.8241-\frac{{\left(0.8241\right)}^3}{3}\right]\\ A=1.0947k\end{array}$

But,

  $A=2$

  $\begin{array}{l}1.0947k=2\\ k=\frac{2}{1.0947}\\ k\approx 1.8269\end{array}$