Finding the area enclosed by the graphs of: $y=x\left(x-4\right)^2$ ,$y=4x-x^2$.

Question

Find the area enclosed by the graphs of: $y=x\left(x-4\right)^2$ , $y=4x-x^2$.

My answer was $\frac{7}{12}+\frac{45}{4}$, but apparently this is wrong: the right answer is $\frac{37}{2}$ according to the textbook.

Intersections occur where $x(x - 4)^2 = x(4 - x)$.

So $x(x - 4)(x - 4 + 1) = 0$ which has solutions at $x = 0$, $x = 3$ and $x = 4$.

The difference equation is $y = x(x - 4)^2 - (4x + x^2) = x^3 - 9x^2 + 12x$.

\begin{align} \text{Enclosed area} & = \int_0^4 \lvert x^3 - 9x^2 + 12x \rvert \, dx \\ & = \int_0^3 x^3 - 9x^2 + 12x \, dx + \int_3^4 -x^3 + 9x^2 - 12x \, dx \\ & = \left[ \frac14 x^4 - 3x^3 + 6x^2 \right]_0^3 + \left[ -\frac14 x^4 + 3x^3 - 6x^2 \right]_3^4 \\ & = \left( \frac{81}4 - 81 + 54 \right) - \left( 0 \right) + \left( -64 + 192 - 96 \right) - \left( -\frac{81}4 + 81 - 54 \right) \\ & = -\frac{27}4 - 0 + 32 - \frac{27}4 \\ & = \frac{37}2 \end{align}

It appears that they changed the $-x^2$ to $x^2$ in the latter equation when finding the difference equation. Why did they do that?

Answer 1

You are correct. That is a typo. Furthermore it propagates all the way to the end. Your answer is correct.