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

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

$$x\approx-0.24626617$$

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

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

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

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

$$=0.1240526+35.25+41.25+0.1240526$$

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