let the region r be the area enclosed by the function f(x)=x² and g(x)=2x. write an integral in terms of x…

Explanation:

Step1: Find intersection points

Set $x^{2}=2x$, then $x^{2}-2x = 0$, factoring gives $x(x - 2)=0$. So $x = 0$ and $x = 2$. These are the limits of integration for the integral in terms of $x$.

Step2: Determine the integrand in terms of x

For $0\leq x\leq2$, $2x\geq x^{2}$. The area between two - curves $y = f(x)$ and $y = g(x)$ is $A=\int_{a}^{b}(g(x)-f(x))dx$. Here, $a = 0$, $b = 2$, $g(x)=2x$ and $f(x)=x^{2}$, so the integral in terms of $x$ is $\int_{0}^{2}(2x - x^{2})dx$.

Step3: Rewrite the functions in terms of y

From $y=x^{2}$, we have $x=\sqrt{y}$ (since we are in the first - quadrant part of the region, we take the positive square root). From $y = 2x$, we have $x=\frac{y}{2}$.

Step4: Find the limits of integration in terms of y

When $x = 0$, $y = 0$; when $x = 2$, $y = 4$.

Step5: Determine the integrand in terms of y

For $0\leq y\leq4$, $\sqrt{y}\geq\frac{y}{2}$. The area between two curves $x = h(y)$ and $x = k(y)$ is $A=\int_{c}^{d}(h(y)-k(y))dy$. Here, $c = 0$, $d = 4$, $h(y)=\sqrt{y}$ and $k(y)=\frac{y}{2}$, so the integral in terms of $y$ is $\int_{0}^{4}(\sqrt{y}-\frac{y}{2})dy$.

Answer:

$x_1 = 0$ $x_2 = 2$ $\int_{0}^{2}(2x - x^{2})dx$ $y_1 = 0$ $y_2 = 4$ $\int_{0}^{4}(\sqrt{y}-\frac{y}{2})dy$