What is the integral from 0 to 7 of sqrt(1+9x) ?

The integral from 0 to 7 of sqrt(1+9x) is 1022/27

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integral from 0 to 7 of sqrt(1+9x)

\int_{0}^{7} 1 + 9 x d x

\frac{1022}{27}

Decimal Notation

37.85185 \dots

Solution steps

\int_{0}^{7} 1 + 9 x d x

Apply u-substitution

: \int_{1}^{64} \frac{u}{9} d u

= \int_{1}^{64} \frac{u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{1}^{64} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{64}

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{64}

Compute the boundaries:

\frac{1022}{3}

= \frac{1}{9} \cdot \frac{1022}{3}

Simplify

= \frac{1022}{27}

6 y^{^{'''}} + 3 y^{^{''}} = 0

\frac{3}{( x + 6 )}, at x = - 4

\frac{s}{s ^{2} - 2}

y^{^{'}} = 7 x + y

\frac{d y}{d x} + 3 x y^{2} = 0, y (0) = 1

What is the integral from 0 to 7 of sqrt(1+9x) ?
The integral from 0 to 7 of sqrt(1+9x) is 1022/27