What is the integral from 0 to 4 of (sqrt(x)) ?

The integral from 0 to 4 of (sqrt(x)) is 16/3

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integral from 0 to 4 of (sqrt(x))

\int_{0}^{4} (x) d x

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

\int_{0}^{4} x d x

Apply the Power Rule

: [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{4}

= [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= \frac{16}{3}

x \to 0 lim (4 x cos (x))

y^{^{''}} + 81 y = cos (6 t)

f (x) = 5 x - 1, at x = 2

(1.8) (0.5)

x \to 0 lim (\frac{2}{x ^{3} - x})

What is the integral from 0 to 4 of (sqrt(x)) ?
The integral from 0 to 4 of (sqrt(x)) is 16/3