We are given function $f(x)$ defined for each real number $x$, such that it satisfies:

$$6 + f(x) = 2f(-x) +3x^2(\int_{-1}^{1}f(t) \, dt)$$

We need to solve this for $\int_{-1}^{1}f(x)\,dx$

I'm not very experienced in integrating and doing calculus, so I tried to look in the solutions, but they aren't very helpful. They say that we should try to integrate the whole given expression and we will end up with $12 + A = 2A + 2A$ where $A = \int_{-1}^1 f(x) \, dx$

  • 1
    $\begingroup$ Have you tried integrating both sides of the equality with respect to $x$ with bounds from $- 1$ to $1$? $\endgroup$
    – kingW3
    Oct 25, 2019 at 18:51
  • 1
    $\begingroup$ Integrating both sides and noting that $\int_{-1}^{1}f(x)\ dx=\int_{-1}^{1}f(-x)\ dx$ is probably the fastest method. $\endgroup$ Oct 25, 2019 at 19:16

3 Answers 3


Another way to do the question is first by noticing that $$\int_{-1}^1 f(t) dt = \int_{-1}^1 f(x) dx$$ So, one of the terms is what you have to find.

Let this be $A$. Then, we get $6+f(x) = 2f(-x) + 3Ax^2$

Now, we can see that $x^2$ is even. So, if we replace $x$ with $-x$ in the above equation, that term will not change. Doing so, we get $6+f(-x) = 2f(x) + 3Ax^2$

Now, subtracting the above $2$ equations and solving, we get $f(x) = f(-x)$. Substituting this in the above equation, we get $$f(x) = 6 - 3Ax^2$$

Integrating, we get - $$A = 12 - 2A$$

or $A = 4$.

Side note: We also proved in the process, that $f(x) = 6 - 12x^2$ is the only function that solves your given equation.

  • 2
    $\begingroup$ You should note that your proved more: the only such function is $$f(x)=6-12x^2$$ $\endgroup$
    – N. S.
    Oct 25, 2019 at 19:00
  • $\begingroup$ Good point. Added to the answer. $\endgroup$
    – Ishan Deo
    Oct 25, 2019 at 19:05

And what they say is correct. If $A=\int_{-1}^1f(x)\,\mathrm dx$, then$$\int_{-1}^16+f(x)\,\mathrm dx=12+A\tag1$$and$$\int_{-1}^12f(-x)+3x^2A\,\mathrm dx=4A,\tag2$$since $\int_{-1}^1x^2\,\mathrm dx=2$ and $\int_{-1}^1f(-x)\,\mathrm dx=A$. Since $(1)=(2)$, $A=4$.

  • 3
    $\begingroup$ Actually $$\int_{-1}^1 x^2~dx=2/3$$ $x^2$ is not an odd function. $\endgroup$ Oct 25, 2019 at 18:50
  • $\begingroup$ Oops! Quite right! I've edited my answer. Thank you. $\endgroup$ Oct 25, 2019 at 18:53

$$6 + f(x) = 2f(-x) +3x^2(\int_{-1}^{1}f(t) \, dt)= 2f(-x) +3x^2A\implies $$


Change $x$ to $-x$ and you get $$ f(-x)-2f(x)=3Ax^2-6$$

Therefore $$f(x)-2f(-x) = f(-x)-2f(x)$$ which imlies $$f(x)=f(-x)$$

Plugging in $$f(x)-2f(-x)=3Ax^2-6$$ we get $$f(x)=6-3Ax^2$$

Thus $$A=\int_{-1}^1 f(t)dt = \int_{-1}^16-3At^2 dt =12-2A $$

Thus $3A=12$ which gives us $A=4$


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