# Are these the same? $\int_{-2}^2(2x^2-x)^4$ and $2\int_{0}^2(2x^2-x)^4$

$$\int_{-2}^2(2x^2-x)^4\,\mathrm dx\quad\text {and}\quad2\int_{0}^2(2x^2-x)^4\,\mathrm dx$$ I tried to solve this and got different answers, but the other problems that I did got the same answers. These are the same right? I just used symmetry or I'm wrong with the use of symmetry this time? I got $2710.96$ for the first one and after applying symmetry I got $492.29$.

• I have seen the exact same integral in some other question recently... Commented Dec 9, 2014 at 10:30
• Alright let me check thanks Commented Dec 9, 2014 at 10:30
• math.stackexchange.com/questions/1057316/… Commented Dec 9, 2014 at 10:31
• @myself I saw my mistake here my other question had a even function that's why my answer here is different. Anyways thanks! Commented Dec 9, 2014 at 10:32
• It's not so simple as your other question, the integrand is symmetric, but around $x = 1/4$ (the extremum of the parabola $y = 2x^2 - x$) and not around $x=0$ in this case). Commented Dec 9, 2014 at 10:35

$$I=\int_{-a}^a f(x)dx=\int_{-a}^0f(x)dx+\int_0^af(x)dx$$

Set $x=-z$ in the first integral

to find $$I=2\int_0^af(x)dx$$ if $f(x)=f(-x)$

Here, $f(x)=(2x^2-x)^4\implies f(-x)=(2x^2+x)^4$

So, $f(x)$ is neither even nor odd

• I don't think it's even. Commented Dec 9, 2014 at 10:37
• @Myself, Agreed & rectified Commented Dec 9, 2014 at 10:39
• @labbhattacharjee therefore I can't use symmetry, is that right? Commented Dec 9, 2014 at 10:41
• @Mickey, Not directly, But But $f(x)=(2x^2)^4-\binom41(2x^2)^3x+\binom42(2x^2)^2x^2-\binom43(2x^2)x^3+x^4$ $=\underbrace{16x^8+24x^6+x^4}_{\text{even}}-\underbrace{(32x^7+8x^5)}_{\text{odd}}$ Commented Dec 9, 2014 at 10:45

$1.$ Check if the function $f(x) = (2x^2-x)^4$ is symmetric.

$2.$ So check if $f(x) = f(-x)$. You can notice that the function is not symmetric.

Edit:

Symmetry about $x=0$ means the integral value is same on both sides of y-axis(Imagine the graph $y=x^2$). Hence when the lower limit of integral is made $0$, the area is split into two and hence you multiply $\int_{0}^2(2x^2-x)^4$ by $2$.

We proved above that the given $f(x)$ is not symmetric. Hence $\int_{-2}^2(2x^2-x)^4$ is not same as $2\int_{0}^2(2x^2-x)^4$.

• So the answer is $0$? since it's an odd? Commented Dec 9, 2014 at 10:33
• It isn't odd. In fact its positive everywhere. Commented Dec 9, 2014 at 10:34
• @Mickey It's not odd. For it to be odd $f(-x)=-f(x)$. We just need to check if it is symmetric over y axis, as we are integrating over $x$.
– lsp
Commented Dec 9, 2014 at 10:35
• @Lsp oh you're right. since it changes does this mean I can't use the symmetry anymore? Commented Dec 9, 2014 at 10:37
• @Mickey Symmetry about $x=0$ means the integral value is same on both sides. Hence when the lower limit of integral is made $0$, the area is split into two and hence you multiply the value obtained from integral with lower limit $0$ by $2$.
– lsp
Commented Dec 9, 2014 at 10:41