Evaluate the definite integral \int_{1}^{3} (4x^3 - 6x) \, dx. — JAMB UTME

Evaluate the definite integral $\int_{1}^{3} (4x^3 - 6x) \, dx$ .

Answer: 56

The value of the definite integral is 56.

To evaluate the definite integral

\int_{1}^{3} (4x^3 - 6x) \, dx

, integrate the function term-by-term to obtain

F(x) = x^4 - 3x^2

. Evaluating

F(x)

at the upper bound

x = 3

yields

54

, and evaluating at the lower bound

x = 1

yields

-2

. Subtracting the lower bound value from the upper bound value gives

54 - (-2) = 56

Find the indefinite integral of the function

\int (4x^3 - 6x) \, dx = x^4 - 3x^2

Apply the power rule of integration \int x^n \, dx = \frac{x^{n+1}}{n+1}

Substitute the upper limit x = 3 into the antiderivative F(x)

F(3) = 3^4 - 3(3)^2 = 81 - 27 = 54

Evaluate F(b) for b = 3

Substitute the lower limit x = 1 into the antiderivative F(x)

F(1) = 1^4 - 3(1)^2 = 1 - 3 = -2

Evaluate F(a) for a = 1

Subtract F(a) from F(b)

54 - (-2) = 56

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

Definite Integration using the Fundamental Theorem of Calculus

Question