Evaluate the definite integral: ${\int }_{-2}^3(x^2-3) dx$.

• A. $12.5$
• B. $10.5$
• C. $10.33$
• D. $10.83$

Answer: (B)

Verified step by step guidance

1. Step 1: Recall the formula for evaluating a definite integral: \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \), where \( F(x) \) is the antiderivative of \( f(x) \).
2. Step 2: Find the antiderivative of \( f(x) = x^2 - 3 \). The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \), and the antiderivative of \( -3 \) is \( -3x \). Thus, \( F(x) = \frac{x^3}{3} - 3x \).
3. Step 3: Apply the limits of integration \( a = -2 \) and \( b = 3 \) to the antiderivative \( F(x) \). Compute \( F(3) \) and \( F(-2) \) separately.
4. Step 4: Substitute \( x = 3 \) into \( F(x) \): \( F(3) = \frac{3^3}{3} - 3(3) \). Then substitute \( x = -2 \) into \( F(x) \): \( F(-2) = \frac{(-2)^3}{3} - 3(-2) \).
5. Step 5: Compute the definite integral by subtracting \( F(-2) \) from \( F(3) \): \( \int_{-2}^{3} (x^2 - 3) \, dx = F(3) - F(-2) \).