What is the integral from 0 to 12 of sqrt(1+x/4) ?

The integral from 0 to 12 of sqrt(1+x/4) is 56/3

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integral from 0 to 12 of sqrt(1+x/4)

\int_{0}^{12} 1 + \frac{x}{4} d x

\frac{56}{3}

Decimal Notation

18.66666 \dots

Solution steps

\int_{0}^{12} 1 + \frac{x}{4} d x

Apply u-substitution

: \int_{1}^{4} u \cdot 4 d u

= \int_{1}^{4} u \cdot 4 d u

Take the constant out:

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

= 4 \cdot \int_{1}^{4} u d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{14}{3}

= 4 \cdot \frac{14}{3}

Simplify

= \frac{56}{3}

y = 14 (sin (x) + cos (x))

n = 0 \sum \infty e^{3 - 2 n}

x \to 0 lim (\frac{a ^{2 x} - 1}{x})

x \to 7 lim (\frac{1}{x ^{2} ( x + 7 )})

\frac{d}{d x} (a cos (2 x))

What is the integral from 0 to 12 of sqrt(1+x/4) ?
The integral from 0 to 12 of sqrt(1+x/4) is 56/3