What is the integral from 0 to 1 of sqrt(1+144x) ?

The integral from 0 to 1 of sqrt(1+144x) is (145sqrt(145)-1)/(216)

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integral from 0 to 1 of sqrt(1+144x)

\int_{0}^{1} 1 + 144 x d x

\frac{145 145 - 1}{216}

Decimal Notation

8.07884 \dots

Solution steps

\int_{0}^{1} 1 + 144 x d x

Apply u-substitution

: \int_{1}^{145} \frac{u}{144} d u

= \int_{1}^{145} \frac{u}{144} d u

Take the constant out:

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

= \frac{1}{144} \cdot \int_{1}^{145} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{145}

= \frac{1}{144} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{145}

Compute the boundaries:

\frac{290 145}{3} - \frac{2}{3}

= \frac{1}{144} (\frac{290 145}{3} - \frac{2}{3})

Simplify

= \frac{145 145 - 1}{216}

\frac{d y}{d x} = \frac{x ( y - 2 )}{x ^{2} + 4}

t^{2} t

x \to 0 + lim (x ln (x))

x \to 3 - lim (\frac{8}{x - 3})

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

What is the integral from 0 to 1 of sqrt(1+144x) ?
The integral from 0 to 1 of sqrt(1+144x) is (145sqrt(145)-1)/(216)