Find the area under the curve defined by \( f(x) = 2x + 1 \) between the points \( x = 1 \) and \( x = 3 \).

To find the area under the curve defined by \( f(x) = 2x + 1 \) between \( x = 1 \) and \( x = 3 \), we can use definite integration. Here's a step-by-step guide to solving the problem: ### Step 1: Set Up the Integral The area \( A \) under the curve from \( x = a \) to \( x = b \) is given by: \[ A = \int_{a}^{b} f(x) \, dx \] For this problem, \( f(x) = 2x + 1 \), \( a = 1 \), and \( b = 3 \). Therefore: \[ A = \int_{1}^{3} (2x + 1) \, dx \] ### Step 2: Find the Antiderivative To integrate \( 2x + 1 \), find the antiderivative of each term separately. \[ \int (2x) \, dx = x^2 \quad \text{and} \quad \int 1 \, dx = x \] So, the antiderivative of \( 2x + 1 \) is: \[ F(x) = x^2 + x \] ### Step 3: Evaluate the Antiderivative at the Bounds Apply the Fundamental Theorem of Calculus by evaluating \( F(x) \) at \( x = 3 \) and \( x = 1 \): \[ A = F(3) - F(1) = (3^2 + 3) - (1^2 + 1) \] Calculate each part: \[ F(3) = 9 + 3 = 12 \] \[ F(1) = 1 + 1 = 2 \] ### Step 4: Subtract to Find the Area \[ A = 12 - 2 = 10 \] ### Conclusion The area under the curve \( f(x) = 2x + 1 \) from \( x = 1 \) to \( x = 3 \) is **10 square units**.