QuestionCalculate the area between $y=\sqrt{x}$ and $y=0$ over the interval $[-1,1]$ .

Studdy Solution

STEP 1

Assumptions1. The functions are $y=\sqrt{x}$ and $y=0$ . The interval is $[-1,1]$
3. The area enclosed by the curves is the definite integral of the absolute difference of the two functions over the interval

STEP 2

First, we need to consider the domain of the function $y=\sqrt{x}$ . The square root function is only defined for $x \geq0$ . So, we need to adjust our interval to $[0,1]$ .

STEP 3

Now, we can calculate the area enclosed by the curves by integrating the absolute difference of the two functions over the interval $[0,1]$ . $Area = \int_{0}^{1} |\sqrt{x} -0| dx$

STEP 4

The absolute value of $\sqrt{x} -0$ is just $\sqrt{x}$ , so we can simplify the integral.
$Area = \int_{0}^{1} \sqrt{x} dx$

STEP 5

Now, we need to find the antiderivative of $\sqrt{x}$ , which is $\frac{2}{3}x^{3/2}$ .

STEP 6

We can now evaluate the definite integral.
$Area = \left[\frac{2}{3}x^{3/2}\right]_{0}^{1}$

STEP 7

Plug in the upper limit of the integral, which is1.
$Area = \frac{2}{3}(1)^{3/2} - \left[\frac{2}{3}x^{3/2}\right]_{0}$

STEP 8

Plug in the lower limit of the integral, which is0.
$Area = \frac{2}{3} - \frac{2}{3}(0)^{3/2}$

STEP 9

Calculate the area.
$Area = \frac{2}{3} - = \frac{2}{3}$ The area enclosed by $y=\sqrt{x}$ and $y=$ on the interval $[-,]$ is $\frac{2}{3}$ square units.

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