What is the integral from-4 to 12 of sqrt(z+4) ?

The integral from-4 to 12 of sqrt(z+4) is 128/3

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integral from-4 to 12 of sqrt(z+4)

\int_{- 4}^{12} z + 4 d z

\frac{128}{3}

Decimal Notation

42.66666 \dots

Solution steps

\int_{- 4}^{12} z + 4 d z

Apply u-substitution

: \int_{0}^{16} u d u

= \int_{0}^{16} u d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{128}{3}

= \frac{128}{3}

x \to 4 lim (x^{2} + 7 - x)

x^{2} y^{^{''}} - 16 x y^{^{'}} + 72 y = 0

\frac{d}{d x} (\frac{2 x}{x ^{2} + 1})

f (x) = 6 x - x^{2}

\int \frac{1}{5 - 5 x} d x

What is the integral from-4 to 12 of sqrt(z+4) ?
The integral from-4 to 12 of sqrt(z+4) is 128/3