What is the value of the definite integral \int_{1}^{3} (2x + 1) \, dx? — JAMB UTME

What is the value of the definite integral $\int_{1}^{3} (2x + 1) \, dx$ ?

The antiderivative of

2x + 1

x^2 + x

. Evaluating at the upper boundary

x = 3

gives

3^2 + 3 = 12

, and at the lower boundary

x = 1

gives

1^2 + 1 = 2

. Subtracting the lower bound result from the upper bound result yields

12 - 2 = 10

Find the indefinite integral (antiderivative) of

2x + 1

\int (2x + 1) \, dx = x^2 + x

Apply the power rule of integration:

\int x^n dx = \frac{x^{n+1}}{n+1}

Evaluate the antiderivative at the upper limit

x = 3

3^2 + 3 = 9 + 3 = 12

Substitute

x = 3

into

x^2 + x

Evaluate the antiderivative at the lower limit

x = 1

1^2 + 1 = 1 + 1 = 2

Substitute

x = 1

into

x^2 + x

Subtract the lower limit value from the upper limit value

12 - 2 = 10

According to the Fundamental Theorem of Calculus:

\int_{a}^{b} f(x)dx = F(b) - F(a)

Definite Integrals of Linear Polynomials

Question