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

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

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

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

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

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

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}

\int_{x}^{y} (1 + y^{2}) d y

\int_{0}^{4} \frac{x ^{2}}{x + 4} d x

\int_{1}^{2} 4 x^{3} e^{x^{4}} d x

\int_{0}^{r} π (r^{2} - x^{2}) d x

\int_{0}^{2 π} 1 cos^{2} (x) d x

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