12. (3 points) Evaluate the net area between the function $$ \sqrt{x} $$ and the $$ x $$-axis within the interval $$ [1,4] $$. $$ \begin{array}{ll} ext { (A) } 1 & ext { (B) } 3 \ ext { (C) } \frac{14}{3} & ext { (D) } \frac{18}{3}\end{array} $$

Question

12. (3 points) Evaluate the net area between the function $$ \sqrt{x} $$ and the $$ x $$-axis within the interval $$ [1,4] $$. $$ \begin{array}{ll}	ext { (A) } 1 & 	ext { (B) } 3 \ 	ext { (C) } \frac{14}{3} & 	ext { (D) } \frac{18}{3}\end{array} $$

UpStudy AI · Accepted Answer

To evaluate the net area between the function $$ \sqrt{x} $$ and the $$ x $$-axis within the interval $$ [1, 4] $$, we need to compute the definite integral of $$ \sqrt{x} $$ from 1 to 4.

The integral we need to evaluate is:

$$
\int_{1}^{4} \sqrt{x} \, dx
$$

First, we find the antiderivative of $$ \sqrt{x} $$. Recall that $$ \sqrt{x} = x^{1/2} $$. The antiderivative is given by:

$$
\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2}
$$

Now we can evaluate the definite integral:

$$
\int_{1}^{4} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{4}
$$

Calculating the upper limit:

$$
\frac{2}{3} (4)^{3/2} = \frac{2}{3} (8) = \frac{16}{3}
$$

Calculating the lower limit:

$$
\frac{2}{3} (1)^{3/2} = \frac{2}{3} (1) = \frac{2}{3}
$$

Now, we subtract the lower limit from the upper limit:

$$
\int_{1}^{4} \sqrt{x} \, dx = \frac{16}{3} - \frac{2}{3} = \frac{16 - 2}{3} = \frac{14}{3}
$$

Thus, the net area between the function $$ \sqrt{x} $$ and the $$ x $$-axis within the interval $$ [1, 4] $$ is:

$$
\boxed{\frac{14}{3}}
$$
The net area is $$ \frac{14}{3} $$. Therefore, the correct answer is (C) $$ \frac{14}{3} $$.

(3 points) Evaluate the net area between the function and the -axis within the interval .

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Calculus Questions