Question

Evaluate $ \int_{0}^{1} (3x^2 + 2x) \, dx $.

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

To evaluate a definite integral, find the antiderivative, apply the limits, and compute the difference: \( F(b) - F(a) \).

Answer 1

The Correct Option is A

Step 1: Write down the integral We want to evaluate the definite integral: \[ \int_{0}^{1} (3x^2 + 2x) \, dx \] Step 2: Use the property of integrals The integral of a sum is the sum of the integrals: \[ \int (3x^2 + 2x) \, dx = \int 3x^2 \, dx + \int 2x \, dx \] Step 3: Find the antiderivative of each term Recall the power rule for integration: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] - For \(\int 3x^2 \, dx\): \[ 3 \int x^2 \, dx = 3 \times \frac{x^{3}}{3} = x^3 \] - For \(\int 2x \, dx\): \[ 2 \int x \, dx = 2 \times \frac{x^{2}}{2} = x^2 \] Step 4: Combine the antiderivatives \[ \int (3x^2 + 2x) \, dx = x^3 + x^2 + C \] Step 5: Apply the limits of integration (from 0 to 1) Use the Fundamental Theorem of Calculus which states: \[ \int_a^b f(x) \, dx = F(b) - F(a) \] where \(F(x)\) is the antiderivative of \(f(x)\). Here, \[ F(x) = x^3 + x^2 \] Calculate \(F(1)\): \[ F(1) = 1^3 + 1^2 = 1 + 1 = 2 \] Calculate \(F(0)\): \[ F(0) = 0^3 + 0^2 = 0 + 0 = 0 \] Step 6: Subtract the values to get the definite integral \[ \int_{0}^{1} (3x^2 + 2x) \, dx = F(1) - F(0) = 2 - 0 = 2 \] Final Answer: \[ \boxed{2} \]