Integration#1

I solved it correctly but just posting the solution from the link where I have confirmed that the previous answer being 0.557 was wrong and correct is 0.232

$\frac 1{k}={\int_{2a}^{\min(1,5a)}(x-a)(x-2a)^2(x-5a)\,dx}$

Now, the indefinite integral is:

$\int (x-a)(x-2a)^2(x-5a)\,dx=\\\frac15x^5-\frac{5}2ax^4+11a^2x^3-22a^3x^2+20a^4x$

The value of $\frac1{k}$ is one polynomial on the range $a\in [0,1/5]$ and another polynomial on $a\in [1/5,\infty)$ .

You definitely shouldn’t need four intervals. It should just be: $\int_{1/10}^{1/5}+\int_{1/5}^{3/2}$

Wolfram Alpha gives me, for

$\frac{1}{k}=\begin{cases} -\frac{189}{10}a^5&a\in[0,1/5]\\ -\frac{1}{10}(2a-1)^3(8a^2-13a+2)&a\in[1/5,\infty) \end{cases}$

and then gives me:

$\int_{1/10}^{1/5} = -\frac{3969}{20000000}\approx -0.0002 = \boxed{0}$ by considering three significant digits only.

and $\int_{1/5}^{3/2}=\frac{1740037}{7500000}\approx 0.2320$

More exact answer = $\boxed{0.2318...}$