What is the value of the definite integral \int_{1}^{3} (4x - 1) \, dx? — JAMB UTME

What is the value of the definite integral $\int_{1}^{3} (4x - 1) \, dx$ ?

The antiderivative of

4x - 1

2x^2 - x

. Evaluating at the upper boundary

x = 3

gives

2(3)^2 - 3 = 15

, and at the lower boundary

x = 1

gives

2(1)^2 - 1 = 1

. Subtracting the lower limit evaluation from the upper limit evaluation yields

15 - 1 = 14

Find the indefinite integral of the integrand

4x - 1

\int (4x - 1) \, dx = 2x^2 - x

Using the power rule of integration,

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

Substitute the upper limit

x = 3

into the antiderivative.

2(3)^2 - (3) = 2(9) - 3 = 18 - 3 = 15

Evaluating

F(b)

for

F(x) = 2x^2 - x

b = 3

Substitute the lower limit

x = 1

into the antiderivative.

2(1)^2 - (1) = 2 - 1 = 1

Evaluating

F(a)

for

F(x) = 2x^2 - x

a = 1

Subtract the lower limit result from the upper limit result.

15 - 1 = 14

By the Fundamental Theorem of Calculus,

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

Evaluation of Definite Integrals using the Fundamental Theorem of Calculus

Estimated Time:1m 0s

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