the curve x = y³ + 8 is graphed. what is the area bounded by the curve, the y - axis, the line y = -1 and…

Explanation:

Step1: Recall the formula for area with respect to y

The area (A) between a curve (x = f(y)), the (y -)axis, and the lines (y = a) and (y = b) is given by (A=\int_{a}^{b}f(y)dy). Here (f(y)=y^{3}+8), (a=- 1), and (b = 2).

Step2: Apply the definite - integral formula

[ \begin{align*} A&=\int_{-1}^{2}(y^{3}+8)dy\ &=\int_{-1}^{2}y^{3}dy+\int_{-1}^{2}8dy \end{align*} ]

Step3: Use the power - rule for integration (\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C(n\neq - 1))

For (\int_{-1}^{2}y^{3}dy=\left[\frac{y^{4}}{4}\right]{-1}^{2}=\frac{2^{4}}{4}-\frac{(-1)^{4}}{4}=\frac{16}{4}-\frac{1}{4}=\frac{15}{4}) For (\int{-1}^{2}8dy=8\left[y\right]_{-1}^{2}=8(2-( - 1))=8\times3 = 24)

Step4: Sum up the results of the two integrals

[ \begin{align*} A&=\frac{15}{4}+24\ &=\frac{15 + 96}{4}\ &=\frac{111}{4}=27.75 \end{align*} ]

Answer:

(27.75)