Question

Given the function \(f(x)=\left\\{\begin{array}{l}x^{2} \text { for } 0 \leq x<1 \\ \sqrt{x} \text { for } 1 \leq x \leq 2\end{array}\right.\) Compute \(\int_{0}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}\).

Short answer

Answer: The value of the integral $\int_{0}^{2} f(x) dx$ for the given piecewise-defined function is $\frac{4\sqrt{2} - 1}{3}$.

Step-by-step solution

Split the integral into two parts

Since the function is defined differently on the intervals \([0,1)\) and \([1,2],\) we can split the integral as follows: \(\int_{0}^{2} f(x) dx = \int_{0}^{1} x^2 dx + \int_{1}^{2} \sqrt{x} dx\)

Evaluate the first integral

To compute the first integral, we evaluate the antiderivative of the function \(x^2\): \(\int x^2 dx =\frac{1}{3}x^3 + C\) Now find the definite integral on the interval \([0, 1]\): \(\int_{0}^{1} x^2 dx = \left[\frac{1}{3}(1)^3 - \frac{1}{3}(0)^3\right]\) \(= \left[\frac{1}{3}(1) - \frac{1}{3}(0)\right] = \frac{1}{3}\)

Evaluate the second integral

For the second integral, we'll first rewrite the function as a power of x, which then we can evaluate the antiderivative: \(\int \sqrt{x} dx = \int x^{\frac{1}{2}} dx = \frac{2}{3} x^{\frac{3}{2}} + C\) Now find the definite integral on the interval \([1, 2]\): \(\int_{1}^{2} \sqrt{x} dx = \left[\frac{2}{3}(2)^{\frac{3}{2}} - \frac{2}{3}(1)^{\frac{3}{2}}\right]\) \(= \left[\frac{2}{3}(2\sqrt{2}) - \frac{2}{3}(1)\right] = \frac{2}{3}(2\sqrt{2} - 1)\)

Add the results of the two integrals

Now we'll combine the results from Steps 2 and 3 to get the final value of the integral: \(\int_{0}^{2} f(x) dx = \int_{0}^{1} x^2 dx + \int_{1}^{2} \sqrt{x} dx\) \(= \frac{1}{3} + \frac{2}{3}(2\sqrt{2} - 1)\) \(= \frac{1}{3} + \frac{4\sqrt{2}}{3} - \frac{2}{3}\) \(= \frac{4\sqrt{2} - 1}{3}\) Therefore, the integral \(\int_{0}^{2} f(x) dx = \frac{4\sqrt{2} - 1}{3}.\)