What is the integral from 4 to 8 of (sqrt(x-4)) ?

The integral from 4 to 8 of (sqrt(x-4)) is 16/3

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integral from 4 to 8 of (sqrt(x-4))

\int_{4}^{8} (x - 4) d x

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{4}^{8} x - 4 d x

Apply u-substitution

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

= \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}

Compute the boundaries:

\frac{16}{3}

= \frac{16}{3}

\int_{0}^{y} e^{x + y} d x

\int_{- 2}^{2} (8 - ∣ x ∣) d x

\int_{- 2}^{0} e^{x} d x

\int_{0}^{8} x^{2} - 8 x d x

\int_{2}^{4} \frac{1 - ( \frac{x}{2} ) ^{2}}{x} d x

What is the integral from 4 to 8 of (sqrt(x-4)) ?
The integral from 4 to 8 of (sqrt(x-4)) is 16/3