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Chapter 5: Problem 51

Find (i) the net area and (ii) the area of the following regions. Graph the function and indicate the region in question. The region bounded by \(y=x^{1 / 2}\) and the \(x\) -axis between \(x=1\) and \(x=4\)

Short Answer

Expert verified
Question: Find the net area and the actual area of the region bounded by the function \(y = x^{1/2}\) and the \(x\)-axis between \(x=1\) and \(x=4\). Answer: The net area and the actual area of the region are both \(\frac{14}{3}\).

Step by step solution

01

Graph the function

First, sketch the function \(y=x^{1/2}\) and mark the region bounded by it and the \(x\)-axis between \(x=1\) and \(x=4\). This will help visualize the problem and make sure the integration is done correctly.
02

Set up the integral for net area

To find the net area of the region we bounded, we need to find the definite integral of the function \(y=x^{1/2}\) over the interval \([1, 4]\). This can be represented as $$ A_{net} = \int_{1}^4 x^{1/2} dx $$
03

Evaluate the integral for net area

Now, we can evaluate the integral by using the power rule for integration. The power rule states that, if \(y = x^n\), then \(\int y dx = \frac{x^{n + 1}}{n + 1} + C\), where \(C\) is the constant of integration. Here, we have: $$ A_{net} = \int_{1}^4 x^{1/2} dx = \left[\frac{2}{3} x^{3/2}\right]_{1}^4 $$ Next, we substitute the limits of integration \(4\) and \(1\): $$ A_{net} = \frac{2}{3}\left(4^{3/2}-1^{3/2}\right) = \frac{2}{3}(8-1)=\frac{2}{3}(7) $$ So, the net area is \(\frac{14}{3}\).
04

Set up the integral for actual area

Since the function \(y = x^{1/2}\) is positive throughout the interval from \([1,4]\), the net area calculated above will also be the actual area. So, we don't need to find the absolute value of the function or integrate it separately. Therefore, the actual area is \(A = \frac{14}{3}\). To summarize: (i) The net area of the region is \(\frac{14}{3}\). (ii) The actual area of the region is \(\frac{14}{3}\).

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