Find the area of the shaded region. WebAssign Plot у 4h 3 y=V x 2 1 X1, 1) X= 27 1 у х х 5 10 15 20 25
The Correct Answer and Explanation is:
It looks like your question is about finding the area of a shaded region on a plot involving the function $y = \sqrt{x}$, with boundaries at $x = 1$ and $x = 27$. I’ll assume that the shaded area is the region under the curve $y = \sqrt{x}$ from $x = 1$ to $x = 27$.
Step 1: Understand the problem
You have a function:
$$
y = \sqrt{x}
$$
and you want to find the area between this curve and the x-axis over the interval $[1, 27]$.
Step 2: Set up the integral
The area under the curve $y = \sqrt{x}$ from $x = 1$ to $x = 27$ is given by the definite integral:
$$
\text{Area} = \int_{1}^{27} \sqrt{x} \, dx
$$
Rewrite the square root as a power:
$$
\sqrt{x} = x^{1/2}
$$
Step 3: Integrate
Use the power rule for integration:
$$
\int x^{n} dx = \frac{x^{n+1}}{n+1} + C
$$
Here, $n = \frac{1}{2}$, so:
$$
\int x^{1/2} dx = \frac{x^{3/2}}{\frac{3}{2}} = \frac{2}{3} x^{3/2} + C
$$
Step 4: Apply the limits of integration
Evaluate from 1 to 27:
$$
\text{Area} = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{27} = \frac{2}{3} \left( 27^{3/2} – 1^{3/2} \right)
$$
Step 5: Simplify powers
Calculate $27^{3/2}$:
- First, rewrite $27^{3/2} = (27^{1/2})^3 = (\sqrt{27})^3$.
- $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$.
- So, $27^{3/2} = (3\sqrt{3})^3 = 3^3 \times (\sqrt{3})^3 = 27 \times 3^{3/2}$.
Calculate $(\sqrt{3})^3 = 3^{1/2 \times 3} = 3^{3/2} = 3 \times \sqrt{3}$.
Therefore,
$$
27^{3/2} = 27 \times 3 \times \sqrt{3} = 81 \sqrt{3}
$$
Calculate $1^{3/2} = 1$.
Step 6: Final area expression
$$
\text{Area} = \frac{2}{3} (81 \sqrt{3} – 1)
$$
Numerically,
- $\sqrt{3} \approx 1.732$,
- So,
$$
81 \times 1.732 \approx 140.292
$$
Thus,
$$
\text{Area} \approx \frac{2}{3} (140.292 – 1) = \frac{2}{3} \times 139.292 = 92.861
$$
Final answer:
$$
\boxed{92.86}
$$
Explanation (300 words):
The shaded area under a curve defined by a function $y = f(x)$ between two points on the x-axis, $a$ and $b$, is found using the definite integral of the function from $a$ to $b$. In this problem, the function is $y = \sqrt{x}$, and the bounds are $x = 1$ and $x = 27$.
Since $\sqrt{x}$ can be rewritten as $x^{1/2}$, the integral becomes:
$$
\int_1^{27} x^{1/2} \, dx
$$
Applying the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, we increase the exponent by 1, resulting in:
$$
\frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2}
$$
Evaluating this integral from 1 to 27 involves calculating $27^{3/2}$ and $1^{3/2}$. The exponent $3/2$ means “square root and then cube,” or equivalently, “cube and then take the square root.” This converts to:
$$
27^{3/2} = ( \sqrt{27} )^3 = (3 \sqrt{3})^3 = 27 \times 3 \times \sqrt{3} = 81 \sqrt{3}
$$
Substituting back, the definite integral evaluates to:
$$
\frac{2}{3} (81 \sqrt{3} – 1)
$$
Approximating $\sqrt{3} \approx 1.732$, the area is approximately 92.86 square units.
This process highlights the usefulness of integration in calculating areas under curves — a fundamental concept in calculus applied extensively in physics, engineering, and other sciences.