What is the integral from 0 to 1 of sqrt(4x+5) ?

The integral from 0 to 1 of sqrt(4x+5) is (27-5sqrt(5))/6

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integral from 0 to 1 of sqrt(4x+5)

\int_{0}^{1} 4 x + 5 d x

\frac{27 - 5 5}{6}

Decimal Notation

2.63661 \dots

Solution steps

\int_{0}^{1} 4 x + 5 d x

Apply u-substitution

: \int_{5}^{9} \frac{u}{4} d u

= \int_{5}^{9} \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_{5}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{9}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{9}

Compute the boundaries:

18 - \frac{10 5}{3}

= \frac{1}{4} (18 - \frac{10 5}{3})

Simplify

= \frac{27 - 5 5}{6}

\int_{1}^{2} \frac{x + 1}{x} d x

\int_{- \frac{π}{3}}^{0} sin (x) d x

\int_{0}^{x} e^{x - t} y d t

\int_{0}^{t} x e^{x} d x

\int_{0}^{1} 1 + t^{2} d t

What is the integral from 0 to 1 of sqrt(4x+5) ?
The integral from 0 to 1 of sqrt(4x+5) is (27-5sqrt(5))/6