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

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

Integrating

3x^2 - 2x

gives the antiderivative

F(x) = x^3 - x^2

. Evaluating

F(3) - F(1)

gives

(27 - 9) - (1 - 1) = 18 - 0 = 18

Find the indefinite integral of the function

\int (3x^2 - 2x) \, dx = x^3 - x^2

Apply the power rule of integration to each term: \int 3x^2 dx = x^3 and \int 2x dx = x^2.

Evaluate the antiderivative at the upper limit x = 3

(3)^3 - (3)^2 = 27 - 9 = 18

Substitute the upper limit into the antiderivative.

Evaluate the antiderivative at the lower limit x = 1

(1)^3 - (1)^2 = 1 - 1 = 0

Substitute the lower limit into the antiderivative.

Subtract the lower limit evaluation from the upper limit evaluation

18 - 0 = 18

By the Fundamental Theorem of Calculus, \int_{a}^{b} f(x) dx = F(b) - F(a).

Fundamental Theorem of Calculus for Definite Polynomial Integrals

Estimated Time:1m 0s

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