Round #1080

Evaluate the definite integral of (4x - 3) from x = 1 to x = 5. Give the exact value.

## Step 1: Identify the integral to be solved The integral given is ∫(4x - 3) dx from x = 1 to x = 5. ## Step 2: Find the antiderivative of the function To solve the integral, we first find the antiderivative of (4x - 3), which is ∫(4x - 3) dx = 2x^2 - 3x + 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) = 2x^2 - 3x. ## Step 4: Evaluate the antiderivative at the limits of integration We evaluate F(x) at x = 5 and x = 1: - F(5) = 2(5)^2 - 3(5) = 2(25) - 15 = 50 - 15 = 35 - F(1) = 2(1)^2 - 3(1) = 2(1) - 3 = 2 - 3 = -1 ## Step 5: Calculate the definite integral The definite integral is F(5) - F(1) = 35 - (-1) = 35 + 1 = 36. The final answer is: $\boxed{36}$

To evaluate the definite integral of (4x - 3) from x = 1 to x = 5, we can use the power rule of integration and the constant multiple rule. First, integrate the function: ∫(4x - 3) dx = 4∫x dx - 3∫1 dx = 4 * (x^(1+1))/(1+1) - 3x + C = 2x^2 - 3x + C Then, apply the Fundamental Theorem of Calculus: ∫_1^5 (4x - 3) dx = [2x^2 - 3x] from 1 to 5 = [2(5)^2 - 3(5)] - [2(1)^2 - 3(1)] = (50 - 15) - (2 - 3) = 35 + 1 = 36 The exact value of the definite integral is 36.