Question

The value of the integral \( \int_0^1 x^2 \, dx \) is:

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

Hint:

To evaluate a definite integral: - First, find the antiderivative of the function. - Then, substitute the upper and lower limits into the antiderivative and subtract the results.

Answer 1

The Correct Option is A

To evaluate the definite integral \( \int_0^1 x^2 \, dx \), we proceed step-by-step: 1. Find the Antiderivative of \( x^2 \): The antiderivative of \( x^2 \) is obtained by applying the power rule of integration. The power rule states that for any function of the form \( x^n \) (where \( n \neq -1 \)): \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] For \( x^2 \), the power rule gives: \[ \int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3} \] This means the antiderivative of \( x^2 \) is \( \frac{x^3}{3} \). 2. Evaluate the Definite Integral: To evaluate the definite integral, we use the Fundamental Theorem of Calculus. According to this theorem, if \( F(x) \) is the antiderivative of \( f(x) \), then: \[ \int_a^b f(x) \, dx = F(b) - F(a) \] Here, the limits of integration are 0 and 1. Thus, we need to evaluate: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 \] Now, we substitute the upper and lower limits of integration: - When \( x = 1 \), \( \frac{1^3}{3} = \frac{1}{3} \), - When \( x = 0 \), \( \frac{0^3}{3} = 0 \). Subtracting these values: \[ \int_0^1 x^2 \, dx = \frac{1}{3} - 0 = \frac{1}{3} \] Thus, the value of the integral is \( \frac{1}{3} \).