What is the integral from 1 to 5 of 2sqrt(x-1) ?

The integral from 1 to 5 of 2sqrt(x-1) is 32/3

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integral from 1 to 5 of 2sqrt(x-1)

\int_{1}^{5} 2 x - 1 d x

\frac{32}{3}

Decimal Notation

10.66666 \dots

Solution steps

\int_{1}^{5} 2 x - 1 d x

Take the constant out:

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

= 2 \cdot \int_{1}^{5} x - 1 d x

Apply u-substitution

: \int_{0}^{4} u d u

= 2 \cdot \int_{0}^{4} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= 2 [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= 2 \cdot \frac{16}{3}

Simplify

= \frac{32}{3}

\int \frac{1}{4 + x} d x

\frac{d y}{d t} = t^{2} y^{3}, y (0) = - 1

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

\frac{d}{d x} (\frac{tan ( x )}{sec ( x )})

\frac{\partial}{\partial x} (arcsec (x + yz))

What is the integral from 1 to 5 of 2sqrt(x-1) ?
The integral from 1 to 5 of 2sqrt(x-1) is 32/3