# How to solve for $\int_{-1}^{1}f(x)\,dx$ in the given experssion

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 $$A = \int_{-1}^1 f(x) \, dx$$

• Have you tried integrating both sides of the equality with respect to $x$ with bounds from $- 1$ to $1$? – kingW3 Oct 25 '19 at 18:51
• Integrating both sides and noting that $\int_{-1}^{1}f(x)\ dx=\int_{-1}^{1}f(-x)\ dx$ is probably the fastest method. – Alexander51413 Oct 25 '19 at 19:16

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.

• You should note that your proved more: the only such function is $$f(x)=6-12x^2$$ – N. S. Oct 25 '19 at 19:00
• Good point. Added to the answer. – Ishan Deo Oct 25 '19 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 $$(1)=(2)$$, $$A=4$$.

• Actually $$\int_{-1}^1 x^2~dx=2/3$$ $x^2$ is not an odd function. – projectilemotion Oct 25 '19 at 18:50
• Oops! Quite right! I've edited my answer. Thank you. – José Carlos Santos Oct 25 '19 at 18:53

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

$$f(x)-2f(-x)=3Ax^2-6$$

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 $$A=4$$