What is the integral from 0 to 16 of sqrt(x) ?

The integral from 0 to 16 of sqrt(x) is 128/3

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integral from 0 to 16 of sqrt(x)

\int_{0}^{16} x d x

\frac{128}{3}

Decimal

42.66666 \dots

Solution steps

\int_{0}^{16} x d x

Apply the Power Rule

: [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{16}

= [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{16}

Compute the boundaries:

\frac{128}{3}

= \frac{128}{3}

\int_{0}^{4} 4 y^{2} - \frac{y ^{4}}{4} d y

x \to a lim (f (x) + g (x))

f (t) = cos (2 t)

x y \frac{d y}{d x} = 1

\int + \frac{x}{x ^{2} + 1} d x

What is the integral from 0 to 16 of sqrt(x) ?
The integral from 0 to 16 of sqrt(x) is 128/3