What is the value of the definite integral \int_{1}^{2} (4x^3 - 3x^2 + 2) \, dx? — JAMB UTME

What is the value of the definite integral $\int_{1}^{2} (4x^3 - 3x^2 + 2) \, dx$ ?

Integrating each term yields

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

. Substituting the upper boundary gives

F(2) = 12

and the lower boundary gives

F(1) = 2

. Calculating

F(2) - F(1)

gives

12 - 2 = 10

, which is the exact value of the definite integral.

Find the indefinite antiderivative of the integrand

4x^3 - 3x^2 + 2

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

Apply the power rule of integration

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

to each term.

Evaluate the antiderivative at the upper limit

x = 2

F(2) = (2)^4 - (2)^3 + 2(2) = 16 - 8 + 4 = 12

Substitute

x = 2

into

F(x)

Evaluate the antiderivative at the lower limit

x = 1

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

Substitute

x = 1

into

F(x)

Subtract the lower limit value from the upper limit value.

F(2) - F(1) = 12 - 2 = 10

Apply the Fundamental Theorem of Calculus:

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

Definite Integration of Polynomial Functions

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