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

Substitute: Let (u = 2x + 2), so that (du = 2dx) or (dx = \frac{du}{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).

Substitute into the Integral:
[ \int_{2}^{4} \sqrt{u} \cdot \frac{1}{2} du = \frac{1}{2} \int_{2}^{4} u^{1/2} du ]
 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} ]
 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.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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