2
$\begingroup$

enter image description here

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 $5a > 1$

enter image description here

$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro Tamaroff
    May 10 '21 at 7:38
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.