What is the integral from 0 to 16 of sqrt(1+3x) ?

The integral from 0 to 16 of sqrt(1+3x) is 76

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integral from 0 to 16 of sqrt(1+3x)

\int_{0}^{16} 1 + 3 x d x

76

Solution steps

\int_{0}^{16} 1 + 3 x d x

Apply u-substitution

: \int_{1}^{49} \frac{u}{3} d u

= \int_{1}^{49} \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}^{49} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{49}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{49}

Compute the boundaries:

228

= \frac{1}{3} \cdot 228

Simplify

= 76

\frac{d}{d x} (ln (x + y - 1))

x^{2} (\frac{d y}{d x}) = y - y x

y = x^{2} - 6 x, (5, - 5)

y^{^{''}} - 2 y^{^{'}} + 10 y = 9 e t csc (3 t)

y = ln (x x^{2} - 9)

What is the integral from 0 to 16 of sqrt(1+3x) ?
The integral from 0 to 16 of sqrt(1+3x) is 76