Respuesta :
The area under a graph is given by the integral of that function, evaluated in the interval of interest:
[tex]\displaystyle \int_{-3}^2 x^2+4\;dx = \left[\dfrac{x^3}{3}+4x\right]_{-3}^2 = \left[\dfrac{2^3}{3}+4\cdot 2\right]-\left[\dfrac{(-3)^3}{3}+4\cdot(-3)\right] = \left[\dfrac{8}{3}+8\right]-\left[-9-12\right] = \dfrac{32}{3}+21 = \dfrac{95}{3}[/tex]
Answer:
[tex](\frac{95}{3})[/tex]
Step-by-step explanation:
Equation that represents curve is
y = x² + 4
and we have to find the area
under the curve in the interval of [-3, 2] will be
area = [tex]\int\limits^2_ {-3} \,x^{2}+4 dx[/tex]
= [tex][\frac{x^{3} }{3}+4x ]^{2} _{-3}[/tex]
= [tex]\frac{1}{3}[x^{3}]^2_{-3} +4[x]^2_-3[/tex]
= [tex]\frac{1}{3}[(2)^3(-3)^3]+4[(2)-(-3)][/tex]
= [tex]\frac{1}{3}[8+27]+4[2+3][/tex]
= [tex]\frac{1}{3}(35)+20[/tex]
= [tex](\frac{95}{3})[/tex]
