question\nfind the area bounded by (f(x)=x^{3}+3x) and (g(x) = 4x) on (xin-1,1).\nsubmit your answer in…

Explanation:

Step1: Determine the upper - lower function

On the interval $[-1,1]$, we need to find which function is above the other. Let $h(x)=g(x)-f(x)=4x-(x^{3}+3x)= - x^{3}+x$. Set $h(x) = 0$, then $-x^{3}+x=-x(x^{2} - 1)=-x(x - 1)(x + 1)=0$. The roots are $x=-1,0,1$. We can test a value in each sub - interval. For $x\in[-1,0]$, let $x=-0.5$, then $h(-0.5)=-(-0.5)^{3}+(-0.5)=0.125 - 0.5=-0.375<0$, so $f(x)\geq g(x)$ on $[-1,0]$. For $x\in[0,1]$, let $x = 0.5$, then $h(0.5)=-(0.5)^{3}+0.5=-0.125 + 0.5 = 0.375>0$, so $g(x)\geq f(x)$ on $[0,1]$.

Step2: Use the area formula

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ on $[a,b]$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx=\int_{-1}^{0}[(x^{3}+3x)-4x]dx+\int_{0}^{1}[4x-(x^{3}+3x)]dx$. Simplify the integrands: The first integral becomes $\int_{-1}^{0}(x^{3}-x)dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]{-1}^{0}=(0 - 0)-(\frac{1}{4}-\frac{1}{2})=\frac{1}{4}$. The second integral becomes $\int{0}^{1}(x - x^{3})dx=\left[\frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{1}=(\frac{1}{2}-\frac{1}{4})-0=\frac{1}{4}$.

Step3: Calculate the total area

$A=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$.

Answer:

$\frac{1}{2}$