What is the integral from-2 to 2 of sqrt(4x+8) ?

The integral from-2 to 2 of sqrt(4x+8) is 32/3

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integral from-2 to 2 of sqrt(4x+8)

\int_{- 2}^{2} 4 x + 8 d x

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

\int_{- 2}^{2} 4 x + 8 d x

Apply u-substitution

: \int_{0}^{16} \frac{u}{4} d u

= \int_{0}^{16} \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{16} u d u

Apply the Power Rule

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

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

Compute the boundaries:

\frac{128}{3}

= \frac{1}{4} \cdot \frac{128}{3}

Simplify

= \frac{32}{3}

\int_{0}^{2} \frac{3}{2} (1 - x)^{2} d x

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\int_{0}^{4} (4 x^{2} + x^{\frac{3}{2}}) d x

\int_{1}^{3} \frac{x}{x + 1} d x

What is the integral from-2 to 2 of sqrt(4x+8) ?
The integral from-2 to 2 of sqrt(4x+8) is 32/3