What is the value of the definite integral \int_{0}^{4} (x + 1) \, dx? — JAMB UTME

What is the value of the definite integral $\int_{0}^{4} (x + 1) \, dx$ ?

Answer: 12

The value of the definite integral is 12.

Integrating

x + 1

yields

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

. Substituting the upper limit

x = 4

gives

\frac{4^2}{2} + 4 = 8 + 4 = 12

. Substituting the lower limit

x = 0

gives

0

. Calculating

12 - 0

gives the final answer of 12.

Find the antiderivative of the integrand

f(x) = x + 1

\int (x + 1) \, dx = \frac{x^2}{2} + x + C

Apply the power rule of integration

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

x^1

and

1 = x^0

Evaluate the antiderivative at the upper bound

x = 4

and lower bound

x = 0

F(4) = \frac{4^2}{2} + 4 = \frac{16}{2} + 4 = 12

, and

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

Substitute the bounds into the antiderivative expression

F(x) = \frac{x^2}{2} + x

Apply the Fundamental Theorem of Calculus

F(b) - F(a)

F(4) - F(0) = 12 - 0 = 12

Subtract the lower bound result from the upper bound result.

Evaluation of definite integrals of simple linear polynomials

The integral

\int_{0}^{4} (x + 1) \, dx

represents the area of a trapezoid under

y = x + 1

from

x = 0

x = 4

. The parallel vertical sides are

h_1 = 1

and

h_2 = 5

, with width

b = 4

. Using the area formula for a trapezoid:

\text{Area} = \frac{1+5}{2} \times 4 = 3 \times 4 = 12

Estimated Time:45s

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