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

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

Answer: 15

The value of the definite integral is 15.

Integrating

x^2 + 2

with respect to

x

yields

\left[\frac{x^3}{3} + 2x\right]_{0}^{3}

. Substituting the upper bound

x = 3

gives

\frac{27}{3} + 6 = 15

. Substituting the lower bound

x = 0

gives

0

. The net value is

15 - 0 = 15

Find the antiderivative of

x^2 + 2

\frac{x^3}{3} + 2x

Apply the power rule of integration to each term.

Substitute the upper limit

x = 3

\frac{3^3}{3} + 2(3) = 9 + 6 = 15

Substitute the lower limit

x = 0

\frac{0^3}{3} + 2(0) = 0

Compute the difference between upper and lower limit values

15 - 0 = 15

Definite Integration of Polynomial Functions

Estimated Time:45s

Question