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

The integral from 0 to 1 of sqrt(1+7x) is (32sqrt(2)-2)/(21)

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

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

\frac{32 2 - 2}{21}

Decimal

2.05975 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{8} \frac{u}{7} d u

= \int_{1}^{8} \frac{u}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int_{1}^{8} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{8}

= \frac{1}{7} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{8}

Compute the boundaries:

\frac{32 2}{3} - \frac{2}{3}

= \frac{1}{7} (\frac{32 2}{3} - \frac{2}{3})

Simplify

= \frac{32 2 - 2}{21}

\frac{d}{d x} (4 x e^{2 x} + 4 x^{2} e^{2 x})

\int 2 x e^{- 3 x^{2}} d x

\frac{\partial}{\partial x} (x - y^{2})

4 x^{- 2}

\frac{\partial}{\partial x} (\frac{y}{y + 4})

What is the integral from 0 to 1 of sqrt(1+7x) ?
The integral from 0 to 1 of sqrt(1+7x) is (32sqrt(2)-2)/(21)