I am trying to integrate a polynomial but I couldn't get the correct answer somehow. I feel like I'm making a mistake when evaluating the integral.
$$\pi\int_{-1}^1{1-2x^2+x^4}dx=[{x-{2x^3\over3}+{x^5\over5}}]^1_{-1}$$ $$=1-\frac23+\frac15+1-\frac23+\frac15$$ $$=2+{2(-2)\over3}+{2(1)\over5}$$ $$={30-26\over15}=\frac4{15}\pi$$
But the integral calculator is giving me: $\frac{16}{15}\pi$
$$\left.\left(x-\frac23x^3+\frac15x^5\right)\right|_{-1}^1=1-(-1)-\frac23(1-(-1))+\frac15(1-(-1))=$$
$$=2-\frac43+\frac25=\frac{30-20+6}{15}=\frac{16}{15}$$
You can avoid a lot of these problems in cases like this if you use the property
$$\int_{-a}^af(x)dx=2\int_0^af(x)dx\;,\;\;\;a>0\;,\;\;f(x)\;\;\text{an even function,}$$
so in this case we'd get
$$2\int_0^1(1-2x^2+x^4)dx=2\left[x-\frac23x^3+\frac15x^5\right]_0^1=$$
$$=2\left(1-\frac23+\frac15\right)=2\frac{15-10+3}{15}=2\frac8{15}=\frac{16}{15}$$