the curve x = -1/3 y^2 + 3y is graphed. what is the area bounded by the curve, the y - axis, the line y = 1…

Explanation:

Step1: Recall the formula for area with respect to y

The area $A$ between a curve $x = f(y)$ and the $y -$axis from $y = a$ to $y = b$ is given by $A=\int_{a}^{b}|f(y)|dy$. Here, $f(y)=-\frac{1}{3}y^{2}+3y$, $a = 1$, and $b = 7$.

Step2: Set up the integral

We have $A=\int_{1}^{7}\left(-\frac{1}{3}y^{2}+3y\right)dy$.

Step3: Integrate term - by - term

The antiderivative of $-\frac{1}{3}y^{2}$ is $-\frac{1}{3}\times\frac{y^{3}}{3}=-\frac{y^{3}}{9}$, and the antiderivative of $3y$ is $3\times\frac{y^{2}}{2}=\frac{3y^{2}}{2}$. So the antiderivative of $-\frac{1}{3}y^{2}+3y$ is $F(y)=-\frac{y^{3}}{9}+\frac{3y^{2}}{2}$.

Step4: Evaluate the definite integral

Using the fundamental theorem of calculus $A = F(7)-F(1)$. $F(7)=-\frac{7^{3}}{9}+\frac{3\times7^{2}}{2}=-\frac{343}{9}+\frac{3\times49}{2}=-\frac{343}{9}+\frac{147}{2}=\frac{-343\times2 + 147\times9}{18}=\frac{-686+1323}{18}=\frac{637}{18}$. $F(1)=-\frac{1^{3}}{9}+\frac{3\times1^{2}}{2}=-\frac{1}{9}+\frac{3}{2}=\frac{-2 + 27}{18}=\frac{25}{18}$. $A=\frac{637}{18}-\frac{25}{18}=\frac{637 - 25}{18}=\frac{612}{18}=34$.

Answer:

34