Evaluate the definite integral \int_{0}^{2} (3x^2 + 4) \, dx. — JAMB UTME

Evaluate the definite integral $\int_{0}^{2} (3x^2 + 4) \, dx$ .

Answer: 16

The value of the definite integral is

16

Integrating

3x^2 + 4

with respect to

x

gives the antiderivative

F(x) = x^3 + 4x

. Evaluating this antiderivative at the upper limit

x = 2

yields

2^3 + 4(2) = 16

, and at the lower limit

x = 0

yields

0^3 + 4(0) = 0

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

16 - 0 = 16

Integrate the polynomial term by term

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

Apply the power rule of integration \int x^n dx = \frac{x^{n+1}}{n+1} and \int k dx = kx.

Apply the fundamental theorem of calculus with limits 0 and 2

[x^3 + 4x]_0^2 = (2^3 + 4(2)) - (0^3 + 4(0)) = 16 - 0 = 16

Evaluate F(b) - F(a) where F(x) is the antiderivative.

Definite Integral Evaluation using the Fundamental Theorem of Calculus

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