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

The integral from-1 to 4 of sqrt(3x+4) is 14

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

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

14

Solution steps

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

Apply u-substitution

: \int_{1}^{16} \frac{u}{3} d u

= \int_{1}^{16} \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_{1}^{16} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{16}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{16}

Compute the boundaries:

42

= \frac{1}{3} \cdot 42

Simplify

= 14

\int_{- 1}^{0} (3 x^{2} - 2 x - \frac{1}{2}) d x

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\int_{0}^{4} x^{2} - 8 x + 16 d x

What is the integral from-1 to 4 of sqrt(3x+4) ?
The integral from-1 to 4 of sqrt(3x+4) is 14