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

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

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

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

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{0} - u d u

= \int_{4}^{0} - u d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{4} - u d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{0}^{4} u d u)

Apply the Power Rule

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

= - (- [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4})

Simplify

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

Compute the boundaries:

\frac{16}{3}

= \frac{16}{3}

f (x) = e^{x}, at x = 1

\frac{d y}{d x} y = 3^{x}

x \to \frac{1}{4} lim (8 x^{3})

\int x (4 x + 7)^{8} d x

\frac{d}{d x} ((x^{2} + 2 x) ln (2 x))

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