This is the question I tried to solve and got answer as wrong 3 times. $\color{blue}{\text{I feel the correct answer should be 0.23 but it says correct is 0.557}}$.
My try: I split the integral at $a=\frac15,\frac12,1$ like $$\left( \int\limits_{1/10}^{1/5}+\int\limits_{1/5}^{1/2} + \int\limits_{1/2}^{1}+\int\limits_{1}^{3/2} \right) \frac1k da$$ and split the given integral accordingly with the domains of the 'x' considering the fact when should be $5a < 1$ or $5a > 1$. Please any help will be appreciated.
I think the one who posted the question, wants us to use the given integral equated to 1, and when i solved that for a particular case when $a<\frac15$ i got $$\frac1k = -\frac{189a^5}{10}$$ I feel we need to complete all those scenarios. I have checked for all scenarios and got 0.23 approx. You can check this integrand if it's useful, when I solved for the case when $5a > 1$
Your answer appears correct.
$$\frac 1{k(a)}={\int_{2a}^{\min(1,5a)}(x-a)(x-2a)^2(x-5a)\,dx}$$
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 $\frac1{k(a)}$ 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
$$\frac{1}{k(a)}=\begin{cases} -\frac{189}{10}a^5&a\in[0,1/5]\\ -\frac{1}{10}(2a-1)^3(8a^2-13a+2)&a\in[1/5,\infty) \end{cases}$$
and then gives me:
$$\int_{1/10}^{1/5} = -\frac{3969}{20000000}\approx -0.0002$$
and $$\int_{1/5}^{3/2}=\frac{1740037}{7500000}\approx 0.2320$$