Answer to Question #276189 in Calculus for khan

Question

Calculate the area under the curve y=x3 +4x+1 from x=-3 to x=3.

Answer 1

$y=0=>x^3+4x+1=0$

$x\approx-0.24626617$

$Area=-\displaystyle\int_{-3}^{-0.24626617}(x^3+4x+1)dx$

$+\displaystyle\int_{-0.24626617}^{3}(x^3+4x+1)dx$

$=-[\dfrac{x^4}{4}+2x^2+x]\begin{matrix} -0.24626617 \\ -3 \end{matrix}$

$+[\dfrac{x^4}{4}+2x^2+x]\begin{matrix} 3\\ -0.24626617 \end{matrix}$

$=0.1240526+35.25+41.25+0.1240526$

$\approx76.748105 ({units}^2)$