What is the integral from 0 to 1 of sqrt(2x+4) ?

The integral from 0 to 1 of sqrt(2x+4) is (2(3sqrt(6)-4))/3

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integral from 0 to 1 of sqrt(2x+4)

\int_{0}^{1} 2 x + 4 d x

\frac{2 ( 3 6 - 4 )}{3}

Decimal

2.23231 \dots

Solution steps

\int_{0}^{1} 2 x + 4 d x

Apply u-substitution

: \int_{4}^{6} \frac{u}{2} d u

= \int_{4}^{6} \frac{u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{4}^{6} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{6}

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{6}

Compute the boundaries:

46 - \frac{16}{3}

= \frac{1}{2} (46 - \frac{16}{3})

Simplify

= \frac{2 ( 3 6 - 4 )}{3}

\int_{\frac{1}{2}}^{\frac{1}{3}} (cos (π x)) d x

\int_{0}^{1} 2 π x 1 + 4 x^{2} d x

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

\int_{0}^{8} \frac{3 x}{1 + x} d x

\int_{0}^{2 π} 3 sin (x) d x

What is the integral from 0 to 1 of sqrt(2x+4) ?
The integral from 0 to 1 of sqrt(2x+4) is (2(3sqrt(6)-4))/3