How do you find the area of the region under the given curve y = (2x + 2)^1/2 on the interval [0,1]?

Question

Emma Aldridge · Accepted Answer

#8/3- 2sqrt2 /3# The area of the region bounded by the curve and the x-axis would be #int_0^1 ydx# = #int_0^1 sqrt(2x+2) dx# = #2/3 sqrt2 (x+1)^(3/2)]_o^1 # =#2/3 sqrt2 (2^(3/2) -1)#= #8/3- 2sqrt2 /3#

Axel Alvarez · Answer

undefined To find the area under the curve $y = \sqrt{2x + 2}$ on the interval $[0,1]$, you need to set up and evaluate an integral. The integral will give you the area under the curve from $x = 0$ to $x = 1$.

The integral you need to evaluate is:

$$
\text{Area} = \int_{0}^{1} \sqrt{2x + 2}\, dx
$$

To solve this integral, you can use a substitution method. Let's solve this step-by-step.

1. **Substitute**: Let $u = 2x + 2$, so that $du = 2dx$ or $dx = \frac{du}{2}$.
   
2. **Change the Limits**: When $x = 0$, $u = 2(0) + 2 = 2$. When $x = 1$, $u = 2(1) + 2 = 4$. So, the new limits of integration are from $u = 2$ to $u = 4$.

3. **Substitute into the Integral**:
   
$$
\int_{2}^{4} \sqrt{u} \cdot \frac{1}{2} du = \frac{1}{2} \int_{2}^{4} u^{1/2} du
$$

4. **Integrate**:
   
Using the power rule for integration, $\int u^{n} du = \frac{u^{n+1}}{n+1} + C$, we find:

$$
\frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_{2}^{4} = \frac{1}{2} \cdot \frac{2}{3} [u^{3/2}]_{2}^{4} = \frac{1}{3} [u^{3/2}]_{2}^{4}
$$

5. **Evaluate the Definite Integral**:
   
Substitute the limits of integration back in to get the area:

$$
\frac{1}{3} [4^{3/2} - 2^{3/2}]
$$

Let's calculate this to get the final answer.The area of the region under the curve $y = \sqrt{2x + 2}$ on the interval $[0,1]$ is approximately $1.724$ square units.