Question

If \( f(x) = \begin{cases} x^2, & 0 \le x<1 \\ \sqrt{x}, & 1 \le x \le 2 \end{cases} \), then \( \int_0^2 f(x) dx = \)

• A. \( \frac{4\sqrt{2} - 1}{3} \)
• B. \( \frac{4\sqrt{2} + 1}{3} \)
• C. \( \frac{4\sqrt{2} - 1}{6} \)
• D. \( \frac{4\sqrt{2} + 1}{6} \)

Hint:

For piecewise defined functions, split the definite integral over the intervals where the function has a consistent definition. Evaluate each integral separately using the power rule for integration: \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \). Remember to apply the limits of integration after finding the antiderivative.

Answer 1

The Correct Option is A

We need to evaluate the definite integral \( \int_0^2 f(x) dx \).
Since the function \( f(x) \) is defined piecewise, we split the integral into two parts: $$ \int_0^2 f(x) dx = \int_0^1 f(x) dx + \int_1^2 f(x) dx $$ For \( 0 \le x<1 \), \( f(x) = x^2 \).
For \( 1 \le x \le 2 \), \( f(x) = \sqrt{x} = x^{1/2} \).
So, the integral becomes: $$ \int_0^2 f(x) dx = \int_0^1 x^2 dx + \int_1^2 x^{1/2} dx $$ Evaluating the first integral: $$ \int_0^1 x^2 dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3} $$ Evaluating the second integral: $$ \int_1^2 x^{1/2} dx = \left[ \frac{x^{1/2 + 1}}{1/2 + 1} \right]_1^2 = \left[ \frac{x^{3/2}}{3/2} \right]_1^2 = \frac{2}{3} \left[ x^{3/2} \right]_1^2 $$ $$ = \frac{2}{3} (2^{3/2} - 1^{3/2}) = \frac{2}{3} (2 \sqrt{2} - 1) = \frac{4\sqrt{2} - 2}{3} $$ Now, add the results of the two integrals: $$ \int_0^2 f(x) dx = \frac{1}{3} + \frac{4\sqrt{2} - 2}{3} = \frac{1 + 4\sqrt{2} - 2}{3} = \frac{4\sqrt{2} - 1}{3} $$