Question

For the curve \(y=\sqrt{x},\) between \(x=0\) and \(x=2,\) find: The area under the curve.

Short answer

\(\frac{4\sqrt{2}}{3}\)

Step-by-step solution

- Identify the integral bounds and function

The problem asks for the area under the curve for the function \(y = \sqrt{x}\) between \(x = 0\) and \(x = 2\). Clearly define the integral bounds [0, 2] and the function \(f(x) = \sqrt{x}\).

- Setup the definite integral

To find the area under the curve, we need to calculate the definite integral of \(\sqrt{x}\) from \(x = 0\) to \(x = 2\). This can be written as \[\int_{0}^{2} \sqrt{x} \, dx\].

- Find the antiderivative

The next step is to find the antiderivative of \(\sqrt{x}\). Recall that \(\sqrt{x} = x^{1/2}\). The antiderivative of \(x^{n}\) is \(\frac{x^{n+1}}{n+1}\). Applying this, the antiderivative of \(x^{1/2}\) is \[\frac{x^{(1/2)+1}}{(1/2)+1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}\].

- Evaluate the definite integral

Now, we evaluate the definite integral using the antiderivative we found. We need to find \(\left. \frac{2}{3} x^{3/2} \right|_{0}^{2}\). Substitute the bounds into the antiderivative: \[\frac{2}{3} (2^{3/2}) - \frac{2}{3} (0^{3/2})\].

- Simplify the expression

Calculate \(2^{3/2}\): \(2^{3/2} = (\sqrt{2})^3 = 2\sqrt{2}\). Therefore, the expression simplifies to \[\frac{2}{3} \cdot 2\sqrt{2} = \frac{4\sqrt{2}}{3}\]. Hence, the area under the curve is \[\frac{4\sqrt{2}}{3}\].

Key concepts

Area Under the Curve

The concept of finding the 'area under the curve' is a fundamental application of definite integrals. When we talk about finding the area under a curve defined by a function, we mean the region enclosed between the graph of the function, the x-axis, and the vertical lines corresponding to the integral bounds. For the given problem, our function is \(y = \sqrt{x}\) and we're interested in the area from \(x = 0\) to \(x = 2\). To find this area, we use the definite integral. If you sketch the curve of \(y = \sqrt{x}\), you can visualize this area as the shaded region beneath the curve and above the x-axis within the given bounds.

Antiderivative

To compute the area under the curve, we need to find the antiderivative of the function. An antiderivative is essentially a function whose derivative gives back the original function. For example, if \(F(x)\) is an antiderivative of \(f(x)\), then \(F'(x) = f(x)\). In our problem, \(f(x) = \sqrt{x}\) which can also be written as \(x^{1/2}\). The antiderivative of \(x^n\) is given by \(\frac{x^{n+1}}{n+1}\). Using this rule, the antiderivative of \(x^{1/2}\) is: \[ \frac{x^{(1/2)+1}}{(1/2)+1} = \frac{x^{3/2}}{3/2} = \frac{2x^{3/2}}{3} \] This antiderivative, \(\frac{2}{3}x^{3/2}\), will help us evaluate the definite integral within the given bounds.

Integral Bounds

Integral bounds are the limits between which you integrate the function. These bounds define the region over which you want to calculate the area under the curve. In the given problem, the bounds are \[0, 2\]. This means we're looking at the area under the graph of \(y = \sqrt{x}\) from \(x = 0\) to \(x = 2\). To evaluate the definite integral, we need to apply these bounds to our antiderivative. This involves calculating the value of the antiderivative at the upper bound, then subtracting the value of the antiderivative at the lower bound. In our problem, we found the antiderivative to be \(\frac{2}{3}x^{3/2}\). By substituting \(x = 2\) and \(x = 0\) into this expression, we get: \[ \frac{2}{3}(2^{3/2}) - \frac{2}{3}(0^{3/2}) = \frac{2}{3} \times 2\sqrt{2} - 0 = \frac{4\sqrt{2}}{3} \] This final value, \(\frac{4\sqrt{2}}{3}\), represents the exact area under the curve between \(x = 0\) and \(x = 2\).