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

What is the value of the definite integral $\int_{0}^{4} \sqrt{x} \, dx$ ?

\frac{16}{3}

Applying the power rule to

x^{1/2}

gives

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

. Substituting the upper limit

x = 4

yields

\frac{2}{3}(4)^{3/2} = \frac{2}{3}(8) = \frac{16}{3}

, and substituting the lower limit

x = 0

yields 0. Subtracting the lower limit result from the upper limit result gives

\frac{16}{3}

Express the integrand with a fractional exponent

\sqrt{x} = x^{1/2}

Rewriting the square root as a fractional exponent allows the application of the power rule of integration.

Find the antiderivative using the power rule of integration

\int x^{1/2} \, dx = \frac{x^{1/2 + 1}}{1/2 + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}

The power rule states that

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

for

n \neq -1

Evaluate the antiderivative at the limits 4 and 0

\left[ \frac{2}{3}x^{3/2} \right]_{0}^{4} = \frac{2}{3}(4)^{3/2} - \frac{2}{3}(0)^{3/2} = \frac{2}{3}(8) - 0 = \frac{16}{3}

By the Fundamental Theorem of Calculus, evaluate

F(b) - F(a)

Definite Integration using the Power Rule

Question