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

The integral from 0 to 3 of sqrt(3x+16) is 122/9

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

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

\frac{122}{9}

Decimal Notation

13.55555 \dots

Solution steps

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

Apply u-substitution

: \int_{16}^{25} \frac{u}{3} d u

= \int_{16}^{25} \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_{16}^{25} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{25}

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{25}

Compute the boundaries:

\frac{122}{3}

= \frac{1}{3} \cdot \frac{122}{3}

Simplify

= \frac{122}{9}

y = ln ∣ sin (x) ∣

\frac{d}{d x} ((x^{2 - x})^{5} (x - 3 x^{2}))

\int \frac{81}{x ^{3} - 9 x ^{2}} d x

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

f (x) = 7 + 10 x

What is the integral from 0 to 3 of sqrt(3x+16) ?
The integral from 0 to 3 of sqrt(3x+16) is 122/9