question find the area bounded by y = x^3 + 2, y = 1, x = -2 and x = 2. submit your answer in fractional…

Explanation:

Step1: Determine the upper - lower functions

First, we need to find where $y = x^{3}+2$ and $y = 1$ intersect. Set $x^{3}+2=1$, then $x^{3}=-1$, so $x = - 1$. For $-2\leqslant x\leqslant - 1$, the upper function is $y = 1$ and the lower function is $y=x^{3}+2$. For $-1\leqslant x\leqslant2$, the upper function is $y = x^{3}+2$ and the lower function is $y = 1$.

Step2: Set up the integral for the area

The area $A$ is given by the sum of two definite - integrals: $A=\int_{-2}^{-1}[(1-(x^{3}+2))]dx+\int_{-1}^{2}[(x^{3}+2 - 1)]dx$. Simplify the integrands: The first integrand is $1-(x^{3}+2)=-x^{3}-1$, and the second integrand is $x^{3}+1$.

Step3: Evaluate the first integral

$\int_{-2}^{-1}(-x^{3}-1)dx=\left[-\frac{x^{4}}{4}-x\right]_{-2}^{-1}$. $=\left(-\frac{(-1)^{4}}{4}-(-1)\right)-\left(-\frac{(-2)^{4}}{4}-(-2)\right)$. $=\left(-\frac{1}{4}+1\right)-\left(-4 + 2\right)$. $=\frac{3}{4}+2=\frac{3 + 8}{4}=\frac{11}{4}$.

Step4: Evaluate the second integral

$\int_{-1}^{2}(x^{3}+1)dx=\left[\frac{x^{4}}{4}+x\right]_{-1}^{2}$. $=\left(\frac{2^{4}}{4}+2\right)-\left(\frac{(-1)^{4}}{4}+(-1)\right)$. $=(4 + 2)-\left(\frac{1}{4}-1\right)$. $=6-\left(\frac{1 - 4}{4}\right)=6+\frac{3}{4}=\frac{24 + 3}{4}=\frac{27}{4}$.

Step5: Calculate the total area

$A=\frac{11}{4}+\frac{27}{4}=\frac{11 + 27}{4}=\frac{38}{4}=\frac{19}{2}$.

Answer:

$\frac{19}{2}$