Calculating Area Between Curves $y=x$ and $y=3\sqrt{x}$

The problem asks for the area of the region enclosed between the curves $y = x$ and $y = 3\sqrt{x}$, bounded by the lines $x = 0$ and $x = 1$.

Step 1: Determine the upper and lower curves

To find the area, we first need to know which function has greater values within the interval $[0, 1]$. Let's test a value, for example, $x = 0.25$:

• For $y = x$, when $x = 0.25$, $y = 0.25$.
• For $y = 3\sqrt{x}$, when $x = 0.25$, $y = 3\sqrt{0.25} = 3 \times 0.5 = 1.5$.

Since $1.5 > 0.25$, the curve $y = 3\sqrt{x}$ is above the curve $y = x$ in the interval $[0, 1]$.

Step 2: Set up the definite integral

The area (A) between two curves $f(x)$ and $g(x)$ from $x=a$ to $x=b$, where $f(x) \geq g(x)$ in the interval, is given by:

$A = \int_{a}^{b} [f(x) - g(x)] dx$

In this case, $f(x) = 3\sqrt{x}$, $g(x) = x$, $a = 0$, and $b = 1$.

The integral for the area is:

$A = \int_{0}^{1} (3\sqrt{x} - x) dx$

Step 3: Evaluate the integral

First, rewrite $3\sqrt{x}$ as $3x^{1/2}$.

$A = \int_{0}^{1} (3x^{1/2} - x) dx$

Now, find the antiderivative:

• The antiderivative of $3x^{1/2}$ is $3 \times \frac{x^{(1/2 + 1)}}{(1/2 + 1)} = 3 \times \frac{x^{3/2}}{3/2} = 3 \times \frac{2}{3} x^{3/2} = 2x^{3/2}$.
• The antiderivative of $x$ is $\frac{x^2}{2}$.

So, the antiderivative of $(3x^{1/2} - x)$ is $2x^{3/2} - \frac{x^2}{2}$.

Step 4: Calculate the definite integral

Evaluate the antiderivative at the limits of integration ($x=1$ and $x=0$):

$A = \left[ 2x^{3/2} - \frac{x^2}{2} \right]_{0}^{1}$

$A = \left( 2(1)^{3/2} - \frac{(1)^2}{2} \right) - \left( 2(0)^{3/2} - \frac{(0)^2}{2} \right)$

$A = \left( 2(1) - \frac{1}{2} \right) - (0 - 0)$

$A = 2 - \frac{1}{2}$

$A = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$

$A = 1.5$

Step 5: Round the result

The area is exactly $1.5$. Rounding to one decimal place, the area remains $1.5$.