What is the integral from-2 to 1 of sqrt(4x+8) ?

The integral from-2 to 1 of sqrt(4x+8) is 4sqrt(3)

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integral from-2 to 1 of sqrt(4x+8)

\int_{- 2}^{1} 4 x + 8 d x

43

Decimal

6.92820 \dots

Solution steps

\int_{- 2}^{1} 4 x + 8 d x

Apply u-substitution

: \int_{0}^{12} \frac{u}{4} d u

= \int_{0}^{12} \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}^{12} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{12}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{12}

Compute the boundaries:

163

= \frac{1}{4} \cdot 163

Simplify

= 43

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

\int_{- 3}^{2} - 3 ∣ x ∣ d x

\int_{- 1}^{1} x^{299} d x

\int_{0}^{1} 2 e^{- x} d x

\int_{0}^{3} x^{2} - x d x

What is the integral from-2 to 1 of sqrt(4x+8) ?
The integral from-2 to 1 of sqrt(4x+8) is 4sqrt(3)