Find the area bounded by the curves $f(x)=(x-5)^3$ and $g(x)=x-5$

Question

Section 5.5

Find the area bounded by the curves $f(x)=(x-5)^3$ and $g(x)=x-5$

Can somebody verify this solution for me? Thanks!

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First we need to find where these curves intersect to figure out the bounds of integration. So we need to solve

$(x-5)^3=x-5$

Dividing both sides by $x-5$ leads us to:

$(x-5)^2=1$

$=x^2-10x+25=1$

$= x^2-10x+24=0$

$= (x-6)(x-4)=0$

and so $x=6,4$. However, we can see from the picture and also from the equation that these curves also intersect when $x=5$ (since $(5-5)^3$ and $(5-5)$ both equal $0$.

Now here is what makes this problem so challenging: From $x=4$ to $x=5$ we have that $f(x)>g(x)$, but from $x=5$ to $x=6$ we have that $g(x)>f(x)$. Therefore we need to calculate TWO integrals to find the total area.

So the total area bounded by the curves is:

$\int_5^6 g(x)-f(x)dx + \int_4^5 f(x)-g(x)dx$

$= \int_5^6 (x-5)-(x-5)^3dx + \int_4^5 (x-5)^3-(x-5)dx$

$= \int_5^6 (x-5)-(x-5)^3dx + \int_4^5 (x-5)^3-(x-5)dx$

Noting that $(x-5)^3=x^3-15x^2+75x-125$ we see that the next step is:

$= \int_5^6 (x-5)-(x^3-15x^2+75x-125)dx + \int_4^5 (x^3-15x^2+75x-125)-(x-5)dx$

$= \int_5^6 x-5-x^3+15x^2-75x+125dx + \int_4^5 x^3-15x^2+75x-125-x+5dx$

$= \int_5^6 -x^3+15x^2-74x+120dx + \int_4^5 x^3-15x^2+74x-120dx$

$= (\frac{-x^4}{4}+\frac{15x^3}{3}-\frac{74x^2}{2}+120x|_5^6) + (\frac{x^4}{4}-\frac{15x^3}{3}+\frac{74x^2}{2}-120x|_4^5)$

$=\frac{1}{2}$

Answer 1

Yes. (Solution verification is kind of funny. What else am I supposed to say?)

Answer 2

Yes we can.

We first note that the area asked for is that between a cubic and a straight line, both crossing zero at $x=5$. So we change variables (move the coordinate system) to $y=x-5$ such that $f(y)=y^3$ and $g(y)=y$. We now integrate from $y=-1$ to $y=1$ (instead of $x$ between $4$ and $6$.

We next note that the integral is twice the integral from $y=0$ to $y=1$.

We thus obtain $$\int_4^6((x-5)-(x-5)^3)dx =\int_{-1}^1(y-y^3)dy $$ which in turn gives

$$\int_{-1}^1(y-y^3)dy=2\int_{0}^1(y-y^3)dy=2\left[\frac{1}{2}y^2-\frac{1}{4}y^4\right]_{y=0}^{y=1}=2\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{2}.$$