If =1 and =1+i√2 (i=-1), where , are two roots of the equation x^3+ax^2+bx+c=0 (a,b,cR), t…

Question

If $\alpha=1$ and $\beta=1+i\sqrt2$ ($i=\sqrt{-1}$), where $\alpha,\beta$ are two roots of the equation $x^3+ax^2+bx+c=0$ ($a,b,c\in\mathbb{R}$), then $\displaystyle\int_{-1}^{1}(x^3+ax^2+bx+c)\,dx$ is equal to:

Accepted Answer

Since coefficients are real, the third root is $\overline{\beta}=1-i\sqrt2$. So

$$x^3+ax^2+bx+c = (x-1)\big(x-(1+i\sqrt2)\big)\big(x-(1-i\sqrt2)\big) = (x-1)\left[(x-1)^2+2\right]$$

With $y=x-1$: $(x-1)^3+2(x-1) = (x^3-3x^2+3x-1)+2x-2 = x^3-3x^2+5x-3$

$$\int_{-1}^1(x^3-3x^2+5x-3)\,dx = \int_{-1}^1(-3x^2-3)\,dx \quad\text{(odd terms vanish)}$$

$$= -3\left[\frac{x^3}{3}\right]_{-1}^1 - 3(2) = -3\left(\frac23\right)-6 = -2-6 = -8$$

Solutions · 1