## Related questions with answers

Find the area of the region. $f(x)=x\left(x^{2}-3 x+3\right), \ g(x)=x^{2}$

Solutions

Verified### Understanding the exercise:

In part a) we sketched the graph, now we will manually find the area of the region between the graphs. Here are once again the functions and the graph.

$f(x)=x(x^2-3x+3)$

$g(x)=x^2$

Let's find the intersection points between the curves $f(x)=x(x^2-3x+3)$ and $g(x)=x^2$.

$\begin{align*} f(x)&=g(x) \\ x(x^2-3x+3) &= x^2 \\ x(x^2-3x+3)- x^2&=0 \\ x(x^2-4x+3)&=0 \\ x(x-1)(x-3) &=0 \end{align*}$

Hence, $f(x)$ and $g(x)$ intesect each other at $x=0$, $x=1$ and $x=3$. Because of $f(x)>g(x)$ at $x\in [0,1]$ and $f(x)<g(x)$ at $x\in [1,3]$, we have:

$\begin{align*} A&=\int_0^1 [f(x)-g(x)]\ dx + \int_1^3 [g(x)-f(x)]\ dx \\ &=\int_0^1 [x(x^2-3x+3)- x^2]\ dx + \int_1^3 [x^2-x(x^2-3x+3)]\ dx \\ &=\int_0^1 (x^3-4x^2+3x)\ dx-\int_1^3 (x^3-4x^2+3x)\ dx \\ &=\left( \frac{x^4}{4}-\frac{4}{3}x^3+\frac{3}{2}x^2 \right)\Biggr |_0^1-\left( \frac{x^4}{4}-\frac{4}{3}x^3+\frac{3}{2}x^2 \right)\Biggr |_1^3 \\ &=\left( \frac{1}{4}-\frac{4}{3}+\frac{3}{2} \right)-\left( \frac{81}{4}-\frac{4}{3}\cdot 27+\frac{3}{2}\cdot 9 \right)+\left( \frac{1}{4}-\frac{4}{3}+\frac{3}{2} \right) \\ &=-\frac{79}{4} +\frac{100}{3}-\frac{21}{2} \\ &=\frac{-237+400-126}{12} \\ &=\frac{37}{12} \end{align*}$

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