What is the integral from 0 to 1 of sqrt(4x+3) ?

The integral from 0 to 1 of sqrt(4x+3) is (7sqrt(7)-3sqrt(3))/6

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integral from 0 to 1 of sqrt(4x+3)

\int_{0}^{1} 4 x + 3 d x

\frac{7 7 - 3 3}{6}

Decimal Notation

2.22068 \dots

Solution steps

\int_{0}^{1} 4 x + 3 d x

Apply u-substitution

: \int_{3}^{7} \frac{u}{4} d u

= \int_{3}^{7} \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_{3}^{7} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{3}^{7}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{3}^{7}

Compute the boundaries:

\frac{14 7}{3} - 23

= \frac{1}{4} (\frac{14 7}{3} - 23)

Simplify

= \frac{7 7 - 3 3}{6}

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What is the integral from 0 to 1 of sqrt(4x+3) ?
The integral from 0 to 1 of sqrt(4x+3) is (7sqrt(7)-3sqrt(3))/6