Round #1050

Evaluate the definite integral of (2x + 1) from x = 0 to x = 3. Give the exact value.

## Step 1: Identify the integral to be solved The integral given is ∫(2x + 1) dx from x = 0 to x = 3. ## Step 2: Find the antiderivative of the function To solve the integral, we first find the antiderivative of (2x + 1). The antiderivative of 2x is x^2, and the antiderivative of 1 is x. So, the antiderivative of (2x + 1) is x^2 + x + C, where C is the constant of integration. ## Step 3: Apply the Fundamental Theorem of Calculus The Fundamental Theorem of Calculus states that the definite integral of a function f(x) from a to b is equal to F(b) - F(a), where F(x) is the antiderivative of f(x). In this case, F(x) = x^2 + x. ## Step 4: Evaluate the antiderivative at the limits of integration We evaluate F(x) at x = 3 and x = 0. F(3) = (3)^2 + 3 = 9 + 3 = 12. F(0) = (0)^2 + 0 = 0. ## Step 5: Calculate the definite integral The definite integral is F(3) - F(0) = 12 - 0 = 12. The final answer is: $\boxed{12}$

To evaluate the definite integral of (2x + 1) from x = 0 to x = 3, we can use the power rule of integration and the constant multiple rule. First, integrate the function: ∫(2x + 1) dx = ∫2x dx + ∫1 dx = x^2 + x + C. Then, apply the Fundamental Theorem of Calculus: ∫_0^3 (2x + 1) dx = [x^2 + x] from 0 to 3 = (3^2 + 3) - (0^2 + 0) = 9 + 3 = 12. The exact value of the definite integral is 12.