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

The integral from 0 to 1 of sqrt(3x+4) is (14sqrt(7)-16)/9

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

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

\frac{14 7 - 16}{9}

Decimal

2.33783 \dots

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{3} \cdot \int_{4}^{7} u d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{14 7}{3} - \frac{16}{3}

= \frac{1}{3} (\frac{14 7}{3} - \frac{16}{3})

Simplify

= \frac{14 7 - 16}{9}

\int_{0}^{3} ∣ 4 - 2 x ∣ d x

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

\int_{0}^{2} x^{2} - 9 x + 14 d x

\int_{4}^{14} 5 e^{0.02 x} d x

What is the integral from 0 to 1 of sqrt(3x+4) ?
The integral from 0 to 1 of sqrt(3x+4) is (14sqrt(7)-16)/9